Math Core

Module 1.4 · Algebra

Exponents and logarithms

Exponent and logarithm problems on the AMC reward fluency: rewriting everything in a common base, spotting a hidden quadratic, and knowing the log rules well enough to chain them. A logarithm is just an exponent, so every log problem is really an exponent problem in disguise.

Logs are exponents

Definition

Logarithm

For b>0b > 0, b≠1b \ne 1 and x>0x > 0, log⁡bx\log_b x is the exponent you put on bb to get xx:

log⁡bx=y⟺by=x.\log_b x = y \quad\Longleftrightarrow\quad b^y = x.

So log⁡232=5\log_2 32 = 5 and log⁡927=32\log_9 27 = \frac{3}{2} (since 93/2=279^{3/2} = 27). Because blog⁡bx=xb^{\log_b x} = x, a log and an exponential with the same base undo each other.

The graphs of y=2xy = 2^x and y=log⁡2xy = \log_2 x are reflections over y=xy = x. Notice that log⁡2x\log_2 x only exists for x>0x > 0.

y = 2^x and y = log₂ x are inverse functions.Open in grapher →

The rules

Each log rule is an exponent rule read backwards.

RuleWhy
log⁡b(xy)=log⁡bx+log⁡by\log_b (xy) = \log_b x + \log_b ybmbn=bm+nb^m b^n = b^{m + n}
log⁡bxy=log⁡bx−log⁡by\log_b \dfrac{x}{y} = \log_b x - \log_b ybmbn=bm−n\dfrac{b^m}{b^n} = b^{m - n}
log⁡bxk=klog⁡bx\log_b x^k = k \log_b x(bm)k=bkm(b^m)^k = b^{km}
log⁡bx=log⁡cxlog⁡cb\log_b x = \dfrac{\log_c x}{\log_c b}change of base

Change of base gives three consequences that appear constantly on the AMC:

Change-of-base shortcuts

log⁡ab⋅log⁡bc=log⁡ac,log⁡ab=1log⁡ba,log⁡ambn=nmlog⁡ab.\log_a b \cdot \log_b c = \log_a c, \qquad \log_a b = \frac{1}{\log_b a}, \qquad \log_{a^m} b^n = \frac{n}{m}\log_a b.

The first one makes long products of logs telescope. The last one lets you rewrite every log in a problem with the same base.

Worked example: A telescoping product

Evaluate log⁡23⋅log⁡34⋅log⁡45⋯log⁡6364\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{63} 64.

Each base cancels with the previous argument: the whole product is log⁡264=6\log_2 64 = 6.

Worked example: Common base

Solve log⁡2x+log⁡4x+log⁡16x=7\log_2 x + \log_4 x + \log_{16} x = 7.

Rewrite in base 22: log⁡4x=12log⁡2x\log_4 x = \frac{1}{2}\log_2 x and log⁡16x=14log⁡2x\log_{16} x = \frac{1}{4}\log_2 x. So

(1+12+14)log⁡2x=74log⁡2x=7,\left(1 + \frac{1}{2} + \frac{1}{4}\right)\log_2 x = \frac{7}{4}\log_2 x = 7,

giving log⁡2x=4\log_2 x = 4 and x=16x = 16.

Hidden quadratics

An equation with 4x4^x and 2x2^x is a quadratic in t=2xt = 2^x, since 4x=(2x)24^x = (2^x)^2. After solving for tt, remember that t=2xt = 2^x must be positive.

Worked example: A quadratic in disguise

Find the sum of all real solutions of 4x−3⋅2x+1+8=04^x - 3 \cdot 2^{x + 1} + 8 = 0.

Let t=2xt = 2^x. Then 4x=t24^x = t^2 and 2x+1=2t2^{x + 1} = 2t, so t2−6t+8=0t^2 - 6t + 8 = 0 and t=2t = 2 or t=4t = 4. That gives x=1x = 1 or x=2x = 2, with sum 33.

Worked example: Checking the domain

Solve log⁡6x+log⁡6(x−5)=2\log_6 x + \log_6 (x - 5) = 2.

Combine: log⁡6(x(x−5))=2\log_6 \big(x(x - 5)\big) = 2, so x2−5x=36x^2 - 5x = 36, or (x−9)(x+4)=0(x - 9)(x + 4) = 0. But x=−4x = -4 makes log⁡6x\log_6 x undefined, so the only solution is x=9x = 9.

Common mistake

Combining logs can create extraneous solutions. The equation log⁡(x(x−5))=2\log(x(x - 5)) = 2 allows x=−4x = -4, but the original equation doesn't. Always plug solutions back into the original logs.

Tip

To count the digits of a huge power NN, use digits=⌊log⁡10N⌋+1\text{digits} = \lfloor \log_{10} N \rfloor + 1. With log⁡102≈0.30103\log_{10} 2 \approx 0.30103, the number 21002^{100} has ⌊30.103⌋+1=31\lfloor 30.103 \rfloor + 1 = 31 digits.

Practice

Practice 1

Solve 8x=32x−38^x = 32^{x - 3}.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

What is log⁡381−log⁡218+log⁡55\log_3 81 - \log_2 \dfrac{1}{8} + \log_5 \sqrt{5}?

Practice 3

If log⁡ab=3\log_a b = 3, what is log⁡b(a2b)\log_b \left(a^2 b\right)?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Solve log⁡2(log⁡3(log⁡4x))=0\log_2\big(\log_3(\log_4 x)\big) = 0.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Positive numbers a,b,ca, b, c satisfy 2a=32^a = 3, 3b=53^b = 5 and 5c=85^c = 8. What is abcabc?

Practice 6

What is the product of all positive real solutions of xlog⁡2x=16x3x^{\log_2 x} = 16x^3?

Practice 7

Given that log⁡102≈0.30103\log_{10} 2 \approx 0.30103, how many digits does 21002^{100} have?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Positive real numbers xx and yy with x<yx \lt y satisfy log⁡xy+log⁡yx=103\log_x y + \log_y x = \dfrac{10}{3} and xy=144xy = 144. What is x+yx + y?

Real contest practice