Decimal & Irregular Exponents & Roots
Consider the following question:
(x)^(y/z)
Some variation of this question — where the variables are not “easy numbers” — is common in finance, engineering, and STEM fields.
While precise calculation of the answer to many significant digits is typically unreasonable, estimation to at least 2 (and often 3 or 4) significant digits is plausible using basic logarithmic principles — plus a little bit of memorization.
Basic Principles
To estimate (or calculate) logarithms effectively, it helps to know a modest set of log values. These are particularly useful:
log(2) = .301
log(3) = .477
log(7) = .845
(For finance and STEM fields, an estimate of log(e) = .4343 is also helpful.)
The more log figures you know — and the more digits you know for each figure — the more accurate your results will be. However, the estimates above can still be used with impressive effectiveness because of the following core rules:
log(A×B) = log(A) + log(B)
log(A/B) = log(A) − log(B)
This means it is most practical to memorize logarithms for prime numbers, because logs of composite numbers can be computed by factoring and adding logs of the prime factors.
Note: log(5) is skipped above because it can be computed easily:
log(5) = log(10) − log(2) = 1 − .301 = .699
General Rule for Estimating (x)^(y/z)
To estimate (or calculate) x^(y/z) using logarithms, apply:
Estimate log(x).
Multiply your result by (y/z).
Compute 10^(result) (usually by estimation).
In the last step, linear interpolation can be used. More accurate methods exist, but are not as simple (and typically rely on calculus ideas — e.g., quadratic interpolation / second-degree Lagrange interpolation).
Example 1: 216536^.36
Step 1: Estimate log(216536).
216536 is between 10^5 and 10^6, so the integer part is 5. More precisely:
log(216536) = 5 + log(2.16536)
Now estimate log(2.16536).
2.16536 is about 1/6 of the way from 2 to 3, so we can estimate:
log(2.16536) ≈ .33 (or .333 for slightly more accuracy)
(The true value will be slightly higher because the slope of log(x) decreases as x increases.)
So:
log(216536) ≈ 5.333
Step 2: Multiply by .36.
5.333 × .36 ≈ 1.92
Step 3: Estimate 10^1.92.
10^1.92 = 10^1 × 10^.92 = 10 × 10^.92
Now estimate 10^.92.
We can bracket .92 using logs we can compute quickly:
log(8) ≈ 3×log(2) = 3×.301 = .903
log(9) ≈ 2×log(3) = 2×.477 = .954
.92 is about 1/3 of the way from .903 to .954.
So the value should be about 1/3 of the way from 8 to 9, i.e. roughly:
10^.92 ≈ 8.33
Therefore:
10^1.92 ≈ 10 × 8.33 = 83.33
This is correct to about four significant digits.
Example 2: 42^(100/57)
Step 1: Estimate log(42).
42 is between 10^1 and 10^2, so the integer part is 1:
log(42) = 1 + log(4.2)
Now estimate log(4.2).
4.2 is about 1/5 of the way from 4 to 5.
log(4) = 2×log(2) = 2×.301 = .602
log(5) = 1 − log(2) = 1 − .301 = .699
So a quick linear estimate is:
log(4.2) ≈ .622
(Again, the true value will be slightly higher because of decreasing slope.)
So:
log(42) ≈ 1.622
Step 2: Multiply by (100/57).
1.622 × (100/57) = 162.2/57 ≈ 2.845614
Step 3: Estimate 10^2.845614.
10^2.845614 = 10^2 × 10^.845614 = 100 × 10^.845614
Now compare .845614 to a memorized log:
log(7) = .845
So .845614 is just slightly above log(7), and only about 2% of the way toward log(8).
That produces an estimate of about:
702
The actual answer is 704.
Because we used linear interpolation on a logarithmic curve and rough log values, the estimate ends up off by about 0.3%, which is still very strong for a quick mental method.
Example 3: Investments
Michael aggressively invests $100,000 at age 30. He anticipates accruing 10% continuous compound interest annually. At age 30, Chloe invests $500,000 in a much less volatile investment and is confident of accruing 5% continuous compound interest over the long run.
If both estimates are correct, what will be the approximate value of their investments at age 65?
Using the continuous compounding formula P e^(rt), we convert to base-10 logs using:
log(e) ≈ .4343
Michael: r×t = 0.10×35 = 3.5
Chloe: r×t = 0.05×35 = 1.75
Multiply by log(e):
Michael: .4343 × 3.5 ≈ 1.52
Chloe: .4343 × 1.75 ≈ .76
So:
Michael’s growth factor ≈ 10^1.52
Chloe’s growth factor ≈ 10^.76
Now estimate:
10^1.52 = 10 × 10^.52
10^1.5 ≈ 31.6 (a well-known benchmark using logs or square roots)
Also:
10^.75 is between 5 and 6
log(5)=.699 and log(6)=.778, so 10^.75 is between about 5.6 and 5.65
So:
Chloe at 65: $500,000 × 5.6 ≈ $2,825,000
Michael at 65: $100,000 × 31.6 ≈ $3,160,000
Both are relatively accurate estimates.
Decimal and Irregular Exponents and Roots
Calculation League has four complex categories covering decimal and irregular exponents and roots. All of these questions follow the format x^(y/z).
The distinctions are:
Decimal categories: the value y/z is fixed (e.g., 2.9 or .65)
Irregular categories: y and z are randomly generated
Exponent categories: y/z > 1
Root categories: y/z < 1
The following examples illustrate approaches for solving these questions.
Decimal Exponents — Example #1
(50^3.3)/10^5
In this example we are fortunate to receive a round number (50) as the base.
Rewrite:
50^3.3 = 50^3 × 50^.3 (since A^(B+C) = A^B × A^C)
50^3 = 5^3 × 10^3 (since (A×B)^C = A^C × B^C)
So we have:
(5^3 × 10^3 × 50^.3) / 10^5
Cancel 10^3:
5^3 × 50^.3 / 100
Since 5^3 = 125:
1.25 × 50^.3
So the question becomes: how do we estimate 50^.3?
Approach #1 — Logarithms
(For a basic description of how to calculate with logs, refer to the method outlined above.)
log(50) = log(10) + log(5) = 1 + .699 = 1.699
1.699 × .3 = .5097
Now estimate 10^.5097.
Using:
log(3)=.477
log(4)=.602
.5097 lies between them, closer to 3 than 4, giving an estimate between 3.2 and 3.3.
For first-attempt accuracy, the higher answer is often safer because of the decreasing slope of log(x). If given multiple attempts, starting slightly lower can reduce the chance you need to revise both up and down.
So use:
50^.3 ≈ 3.2 (rough estimate)
Then:
1.25 × 3.2 = 4.0
This would be correct for Calculation League purposes.
(The actual answer is 4.042044.)
Approach #2 — Exponent Principles
We can also attempt to compute 50^.3 directly using exponent rules.
Since .3 = 3/10, one option would be:
compute 50^3 = 125000, then take the 10th root
(for example, square root then fifth root — or in another order)
Often it’s more effective to avoid the cube by rewriting using:
A^(B−C) = A^B / A^C
Here:
50^.3 = 50^.5 / 50^.2
This eliminates the cube. (With “50” specifically, the benefit is small — but the principle matters more for other numbers.)
Recall:
without the 50^.3 component, the rest of the expression reduces to 1.25, and the correct final answer is 4.042044.
In Calculation League, acceptable answers are within .05.
So we need:
1.25 × 50^.3 to fall within .05 of 4.042044
i.e., 50^.3 should be between about 3.1936 and 3.2736 for a first-try hit
But for “within five tries” (two guesses up and two guesses down), we only need a rough estimate between about 3.0 and 3.4.
Now apply rough root estimates:
sqrt(50) ≈ 7.07 (for illustration, round to 7)
50^(1/5): since 2^5=32 and 3^5=243, a linear interpolation suggests around 2.1
So:
50^.3 = 50^.5 / 50^.2 ≈ 7 / 2.1 ≈ 10/3 ≈ 3.33
This produces:
1.25 × 3.33 ≈ 4.125
That’s not accurate enough for a guaranteed first-try, but it is close enough that (remembering linear interpolation underestimates 50^.2) you would expect to adjust downward and get the correct answer on a subsequent attempt.
Decimal Exponents — Example #2 (advanced level)
(765^2.6)/10^5
Logarithmic Method
log(7)=.845 and log(8)=.903
Estimate log(7.65) by linear interpolation: ≈ .883
Therefore log(765) ≈ 2.883
Now:
2.6 × 2.883 ≈ 7.5
So the expression becomes:
10^7.5 / 10^5 = 10^2.5
Now estimate 10^2.5:
10^2.5 = 100 × 10^.5 ≈ 100 × 3.162 = 316
Linear interpolation with log(3)=.477 and log(4)=.602 gives an estimate around 318, but the direct observation above is simpler and more accurate.
The actual answer is 314.427 — so 316 should allow a correct answer within five attempts.
Exponent Method
Computing a^.6 is more difficult than a^.3.
Ways to rewrite:
a^.6 = a^.5 × a^.1
(square root first, then compute a/a^.5, then estimate a fifth root, then multiply back)
Or:
a^.6 = a^1 / a^.4
(estimate the fifth root of a, then either square it or divide a by its fifth root twice)
Now compute:
765^2 can be computed as (765+35)(765−35) + 35^2
= (800×730) + 1225
= 584000 + 1225
= 585225
Or use the “ending in 5” trick:
765^2 = 77×76 with 25 appended = 5852|25 = 585225
So:
(765^2.6)/10^5 = (765^2 × 765^.6)/10^5
= (5.85225×10^5 × 765^.6)/10^5
= 5.85225 × 765^.6
At this point, estimating 765^.6 accurately enough is very difficult without advanced fifth-root techniques. This is exactly why, past a certain difficulty level, the logarithmic method becomes the practical default.
Decimal Exponents — Example #3 (expert level)
(3058^3.5)/10^10
In this question we got a relatively friendly exponent (3.5) and a base near 3000.
Logarithmic Method
log(3058) = log(1000) + log(3.058) = 3 + log(3.058)
log(3)=.477 and log(4)=.602 → estimate log(3.058) ≈ .484
so log(3058) ≈ 3.484
Multiply by 3.5:
3.5 × 3.484 = 12.194
So:
10^12.194 / 10^10 = 10^2.194
= 100 × 10^.194
Using linear interpolation, 10^.194 ≈ 1 + 194/302 ≈ 1.643
So estimated answer ≈ 164
This is close, and improving the interpolation would move it into a correct range.
Exponent Technique
Because 3058 is near 3000, we can compute 3058^3 with the Binomial Theorem:
(3000+58)^3
= 3000^3 + 3×3000^2×58 + 3×3000×58^2 + 58^3
Compute:
3000^3 = 27,000,000,000
3×3000^2×58 = 3×9,000,000×58 = 27,000,000×58 = 1,566,000,000
3×3000×58^2 = 9000×3364 = 30,276,000
58^3 = 195,112
Total:
28,596,471,112
Now check what we actually need. Because the exponent is 3.5, we multiply by sqrt(3058):
sqrt(3058) ≈ 55.299 (≈55.3)
The correct final answer is 158.
For Calculation League, a value between 156 and 160 would allow a correct answer within five attempts.
This means we do not need all digits of 3058^3:
If 3058^3 is only 1 significant digit (~30 billion), answer is ~166
If it’s 2 significant digits (~29 billion), answer is ~160
If it’s 3 significant digits (~28.6 billion), answer is ~158 (and we get essentially first-try accuracy)
So: we mainly need the first two binomial terms, and even the second term can be estimated rather than computed perfectly.
Decimal Roots — Example #1
93^.2
In Calculation League, the “roots” versions of decimal/irregular problems are usually simpler than exponent versions, because you avoid the large multiplications.
With up to five submitted answers to get two significant digits, we can use simple milestones:
2^5 = 32
3^5 = 243
(5/2)^5 ≈ 98
So the fifth root of 93 is between 2 and 2.5, and since (5/2)^5 ≈ 98, the answer should be just under 2.5.
Submitting 2.5 or 2.45 without further calculation is reasonable and can be correct on the first try.
Decimal Roots — Example #2 (advanced level)
1815^.65
Logarithmic Method
Because linear interpolation is least accurate in the 1–2 range, numbers starting with 1 or 2 can produce greater error if you start from log(1) and log(2). So rather than starting with log(1.815)+log(1000), we can sometimes start with log(18.15)+log(100).
Option A
Estimate log(1.815) between log(1)=0 and log(2)=.301:
linear estimate gives log(1.815) ≈ .245
so log(1815) ≈ 3.245
Multiply:
3.245 × .65 ≈ 2.109
So answer ≈ 10^2.109 = 100 × 10^.109
Estimate 10^.109 via linear interpolation:
10^.109 ≈ 1 + 109/301 ≈ 1.362
So estimate ≈ 136
This is close, but not quite close enough to guarantee a correct answer without refinement (though it is close enough that small strategy adjustments can make it work).
Option B
Shift to log(18.15)+log(100). Use log(18) and log(20):
log(18) = log(2) + 2×log(3) = .301 + .954 = 1.255
log(20) = 2×log(2) + log(5) = .602 + .699 = 1.301
Even using log(18) directly gives log(1815) ≈ 3 + 1.255 = 4.255 — but since we are using 18.15, we correct via interpolation.
Using the approach as written in your draft:
this produces an estimate around 3.258 for log(1815) (correct is 3.25888)
Then:
3.258 × .65 ≈ 2.118
Now we face the same issue: finishing with a small exponent (.118) amplifies interpolation error. So mirror the earlier fix: interpolate with 1.118 instead.
Use:
log(12) = 2×log(2) + log(3) = .602 + .477 = 1.079
log(14) = log(2) + log(7) = .301 + .845 = 1.146
Interpolate 1.118 between 1.079 and 1.146:
fraction ≈ (1.118−1.079)/(1.146−1.079) = 39/67
so value ≈ 12 + (2×39/67) = 12 + 78/67 = 13.164
So answer ≈ 10 × 13.164 = 131.64, which is correct within .5.
Root Method
Solving a^.65 directly by roots is difficult unless you are comfortable with fifth roots.
One way:
rewrite as a^.5 × a^.1 × a^.05
Using reasonable estimates:
1815^.5 ≈ 42.6
fifth root of 42.6 ≈ 2.1 (since 2^5=32 and 3^5=243)
sqrt(2.1) ≈ 1.45
Then:
42.6 × 2.1 × 1.45 ≈ 130
This is close enough to submit a correct answer — but it is not reliable unless you are very comfortable with fifth roots and with adjustment strategies.
(As you note in your draft: choosing 2.1 instead of 2.05 is influenced by the fact that linear interpolation underestimates, but rounding up can make it harder to know whether your final estimate is an under- or overestimate. In this case it works, but it is not always clear.)
Decimal Roots — Example #3 (expert level)
276^.55
Logarithmic Method
log(2.76) ≈ .435 (linear interpolation)
so log(276) ≈ 2.435
Multiply:
2.435 × .55 ≈ 1.339
Then:
10^1.339 = 10 × 10^.339
Estimate 10^.339 between 2 and 3:
log(2)=.301
log(3)=.477
So 10^.339 ≈ 2.16
Final estimate:
21.6
This is within .5 of the exact answer.
Root Method
Apply a similar idea as above, but note we no longer need to multiply by a^.1, so we can reorder roots.
276^.5 ≈ 16.6
16.6^.5 ≈ 4.07
Now we need 4.07^.05 (i.e., fifth root of sqrt(sqrt(276))) to multiply by 16.6, but we can estimate 4.07^.2 first:
To get first-try accuracy, we’d need 4.07^.2 between 1.3 and 1.35. For five attempts, we only need it between about 1.18 and 1.47.
Estimate milestones:
1.5^2 = 2.25; 2.25^2 > 4.07 → so 4.07^.2 < 1.5
1.4^2 = 1.96; 1.96^2 ≈ 3.84; 3.84×1.4 > 4.07 → so 4.07^.2 < 1.4
1.3^2 = 1.69; 1.69^2 ≈ 2.85; 2.85×1.3 < 4.07 → so 4.07^.2 > 1.3
So it’s between 1.3 and 1.35. Start with 1.3 and guess up if needed:
16.6 × 1.3 = 21.6
Same result as the log method.
Irregular Exponents — Example #1
2^(41/6)/10^8
On irregular exponents, the questions include a denominator to keep the number of significant digits reasonable. Usually the denominator is 10^8, but not always.
Logarithmic Method
log(2)=.301
.301 × (41/6) ≈ 2.057
So:
2^(41/6)/10^8 = 10^2.057 / 10^8 = 10^.057 / 10^6
Using linear interpolation for 10^.057 is inaccurate, so instead rewrite as:
10^1.057 / 10^7
Now estimate 10^1.057 ≈ 10 × 10^.057.
Using only memorized values, you then proceed as written in your draft:
log(12)=2×log(2)+log(3)=1.079
(57/79) scaling leads to 10 × (10 + 114/79) ≈ 114.4
Then adjust for the denominator:
estimated result ≈ .000001144
Root Method
Rewrite:
2^(41/6) = 2^6 × 2^(5/6)
So:
2^6 = 64
need 32^(1/6)
Compute 6th root via cube root then square root (or vice versa):
cube root of 32 ≈ 3.175
sqrt(3.175) ≈ 1.782
Then:
64 × 1.782 ≈ 114.05
Even 1.8 would likely be sufficient.
Irregular Exponents — Example #2 (advanced level)
4.3^(188/93)/10^8
Logarithmic Method
log(4)=.602; log(5)=.699
interpolate log(4.3) ≈ .631
Multiply:
.631 × (188/93) ≈ 1.275
So:
10^1.275 / 10^8 = 10^.275 / 10^7
Estimate 10^.275:
275/301 ≈ .914
so 10^.275 ≈ 1.914
So estimated answer:
.000000191
(correct to an extra digit)
Root Method
Rewrite exponent as integer + remainder:
4.3^(188/93) = 4.3^2 × 4.3^(2/93)
4.3^2 = 18.49
Estimating 4.3^(2/93) precisely is difficult, but since only two significant digits are needed, rounding 18.49 to 19 and guessing upward if needed is a practical competition approach.
Irregular Exponents — Example #3 (expert level)
3.73^(1244/946)/10^8
Logarithmic Method
log(3)=.477; log(4)=.602
interpolate log(3.73) ≈ .568
Multiply:
.568 × (1244/946) ≈ .747
So:
10^.747 / 10^8
Estimate 10^.747:
log(5)=.699 and log(6)=.778 → 10^.747 is between 5 and 6
gives about 5.61
So estimated answer:
.000000056
Root Method
Rewrite exponent as integer + remainder:
3.73^1 × 3.73^(298/946)
Since only two significant digits are needed, estimating cube root of 3.73 (~1.55) can be sufficient in this context:
rounding to 1.5 gives the answer to two significant digits.
Irregular Roots — Example #1
45^(4/948)
Irregular roots follow the same structure as irregular exponents, except the exponent is less than 1.
In some randomly generated irregular-root questions, the value converges close enough to 1 that calculation is unnecessary.
Here:
the 237th root of 45 is barely more than 1
if you are confident the answer lies between 1 and 1.45, Calculation League does not require heavy computation
So here an answer of 1 is appropriate; 1.1 is already an unreasonably high estimate for the 237th root of 45.
Irregular Roots — Example #2 (advanced level)
875835^(2/19)
Logarithmic Method
Rewrite:
(8.75835×10^5)^(2/19)
Estimate log(8.75835):
log(8)=.903; log(9)=.954
interpolate log(8.75835) ≈ .942
So log(base) ≈ 5.942.
Multiply:
5.942 × (2/19) ≈ .625
So:
10^.625
Compare:
log(4)=.602
log(5)=.699
Interpolate gives:
about 4.24 (or 4.2), which matches the correct answer.
Root Method
This is more “art” and depends on target accuracy.
2/19 is between 1/10 and 1/9
so the answer is between the 10th and 9th roots
Estimate 9th root via cube root twice:
875835^(1/3) is between 95 and 96
cube root of 95 is about 4.5
→ so 9th root is about 4.5
Estimate 10th root via square root then fifth root:
sqrt(875835) is between 935 and 936
935^.2 is around 3.9 (since 3^5=243 and 4^5=1024)
Either estimate is enough to submit a correct answer within five attempts, and averaging them supports 4.2, which is correct.
Irregular Roots — Example #3 (expert level)
48250228^(56/914)
Logarithmic Method
Rewrite:
(4.8250228×10^7)^(56/914)
Estimate log(4.8250228):
log(4)=.602; log(5)=.699
interpolate ≈ .682
So log(base) ≈ 7.682.
Multiply:
7.682 × (56/914) ≈ .471
So:
10^.471
Since log(3)=.477, this is close to 3:
start with 3.0 (and note it should be slightly less)
Correct answer is 2.96. A single interpolation step yields ~2.97.
Root Method
We want a value between the 16th and 17th root of an eight-digit number.
Estimating the 16th root (four square roots) is often enough:
48250228^.5 ≈ 6946
6946^.5 ≈ 83
83^.5 ≈ 9.1 (or 9 is enough)
9.1^.5 ≈ 3