Exponents, Exponentials and Logarithms · Lesson 5

Solving exponential and log equations

Match the bases when you can. If both sides are powers of one number, rewrite them and set the exponents equal. 25x=125x−125^x = 125^{x-1} becomes 52x=53x−35^{2x} = 5^{3x-3}, so 2x=3x−32x = 3x - 3 and x=3x = 3.

Take logs when you can't. For 3x=203^x = 20, take the log of both sides: xlog⁡3=log⁡20x\log 3 = \log 20, so x=log⁡20log⁡3x = \frac{\log 20}{\log 3}, which is log⁡320\log_3 20. Isolate the power first: in 4⋅3x=804 \cdot 3^x = 80, divide by 4 before taking logs.

Log equations: condense, then rewrite as a power. log⁡3x+log⁡3(x−6)=3\log_3 x + \log_3(x - 6) = 3 becomes log⁡3[x(x−6)]=3\log_3[x(x - 6)] = 3, so x(x−6)=33=27x(x - 6) = 3^3 = 27. A single log equal to a number turns into an equation without logs.

Check every root in the original equation. Combining logs can create roots that make an original log input zero or negative. Above, x2−6x−27=0x^2 - 6x - 27 = 0 gives x=9x = 9 or x=−3x = -3, and x=−3x = -3 makes log⁡3x\log_3 x undefined, so the only solution is 9.

Worked example Easy

What value of xx satisfies the equation 4x+1=8x4^{x+1} = 8^x?

  1. −2-2
  2. 12\frac{1}{2}
  3. 11
  4. 22Answer

How to solve it

  1. Write both sides as powers of 2: 4=224 = 2^2 and 8=238 = 2^3.
  2. (22)x+1=(23)x(2^2)^{x+1} = (2^3)^x becomes 22x+2=23x2^{2x+2} = 2^{3x}.
  3. Same base, so the exponents are equal: 2x+2=3x2x + 2 = 3x, and x=2x = 2.
  4. Check: 43=644^3 = 64 and 82=648^2 = 64.

Why each choice is right or wrong

  • A. Incorrect. −2-2 is a sign slip when you move 3x3x: 2x+2=3x2x + 2 = 3x gives 2=x2 = x, not x=−2x = -2.
  • B. Incorrect. 12\frac{1}{2} comes from writing 4 as 212^1, which gives x+1=3xx + 1 = 3x. Since 4=224 = 2^2, the left exponent is 2(x+1)2(x + 1).
  • C. Incorrect. 1 multiplies only the xx by 2, writing 22x+12^{2x+1}. The 2 multiplies the whole exponent: 2(x+1)=2x+22(x + 1) = 2x + 2.
  • D. Correct. 22x+2=23x2^{2x+2} = 2^{3x}, so 2x+2=3x2x + 2 = 3x and x=2x = 2. Both sides equal 64.

Worked example Hard

What are all the real solutions of log⁡2(x+2)+log⁡2(x−2)=5\log_2(x + 2) + \log_2(x - 2) = 5?

  1. 66 onlyAnswer
  2. −6-6 only
  3. 33 only
  4. −6-6 and 66

How to solve it

  1. Product rule: log⁡2[(x+2)(x−2)]=5\log_2[(x + 2)(x - 2)] = 5.
  2. Rewrite as a power: (x+2)(x−2)=25=32(x + 2)(x - 2) = 2^5 = 32.
  3. x2−4=32x^2 - 4 = 32, so x2=36x^2 = 36 and x=6x = 6 or x=−6x = -6.
  4. Check: x=−6x = -6 makes x+2=−4x + 2 = -4, a negative log input, so reject it. Only x=6x = 6 works.

Why each choice is right or wrong

  • A. Correct. x=6x = 6 gives log⁡28+log⁡24=3+2=5\log_2 8 + \log_2 4 = 3 + 2 = 5, and x=−6x = -6 makes both inputs negative, so 6 is the only solution.
  • B. Incorrect. −6-6 only keeps the root that fails. At x=−6x = -6 the inputs are −4-4 and −8-8, and the log of a negative number is undefined.
  • C. Incorrect. 3 only sets x2−4=5x^2 - 4 = 5, using 5 as the product instead of as the exponent. The equation means (x+2)(x−2)=25(x + 2)(x - 2) = 2^5.
  • D. Incorrect. −6-6 and 66 both solve x2=36x^2 = 36, but log inputs must be positive. Check each root in the original equation and reject x=−6x = -6.

Trap Keeping an extraneous root

After you combine logs, the quadratic you solve allows inputs the original equation does not. A choice listing both roots is waiting for anyone who skips the check. Put each root back into every original log: each input must be positive.

Trap Comparing exponents before the bases match

4x=8x−14^x = 8^{x-1} does not mean x=x−1x = x - 1. Exponents can be set equal only after both sides share a base: 22x=23x−32^{2x} = 2^{3x-3}, so 2x=3x−32x = 3x - 3 and x=3x = 3.

Calculator When a calculator is faster

Graphing is a strong check here. In the built-in Desmos calculator if you test online, graph each side as its own equation, such as y = log_2(x + 2) + log_2(x - 2) and y = 5, and look for the intersection. A log graph exists only where its input is positive, so an extraneous root simply doesn't appear.

On paper, your calculator can test each choice by substitution, and its log key handles answers such as log⁡12log⁡5\frac{\log 12}{\log 5}.