Precalc
Section 4.6

Exponential & Logarithmic Equations

This is the most tested topic from Chapter 4 on the final. Every technique from sections 4.1–4.5 — exponent rules, log properties, change of base — comes together here.

⚡ Quick Summary

Solving exponential and log equations is like unlocking a door — you just need the right key. For exponential equations (where the variable is in the exponent), ask yourself one question: can I rewrite both sides with the same base? If 8x=328^x = 32, you can rewrite as 23x=252^{3x} = 2^5 and just set 3x=53x = 5. If the bases don't match (like 5x=175^x = 17), take ln⁡\ln of both sides to bring the exponent down where you can work with it.

Same-Base Strategy
bf(x)=bg(x)  ⟹  f(x)=g(x)b^{f(x)} = b^{g(x)} \implies f(x) = g(x)

Only works when both sides share the same base

Take-Log-of-Both-Sides Strategy
ax=c  ⟹  x=ln⁡cln⁡aa^x = c \implies x = \frac{\ln c}{\ln a}

Use when bases cannot be matched

📋 Before You Start

Tap any item if you need a refresher:

Plain-English version

Solving exponential and log equations is like unlocking a door — you just need the right key. For exponential equations (where the variable is in the exponent), ask yourself one question: can I rewrite both sides with the same base? If 8x=328^x = 32, you can rewrite as 23x=252^{3x} = 2^5 and just set 3x=53x = 5. If the bases don't match (like 5x=175^x = 17), take ln⁡\ln of both sides to bring the exponent down where you can work with it.

For log equations, the process goes in reverse. Your goal is to combine everything into one single log on one side (using the product and quotient rules from section 4-5), then convert to exponential form to get rid of the log entirely. For example, log⁡2(x)+log⁡2(x−3)=2\log_2(x) + \log_2(x-3) = 2 becomes log⁡2(x(x−3))=2\log_2(x(x-3)) = 2, which means x(x−3)=4x(x-3) = 4. Now it's just a regular equation you can solve.

Here's the catch that loses people points: extraneous solutions. When you solve a log equation, the algebra might give you answers that don't actually work — specifically, answers that make you take the log of zero or a negative number (which is impossible). You *must* plug each answer back into the original equation and check. If it breaks a log, throw it out.

Think of it as two paths: same-base matching is the express lane (faster, no decimals), and taking ln⁡\ln of both sides is the scenic route (works every time but gives decimals). For log equations, there's only one path: condense, convert, solve, and always check your answers.

Key Formulas

Same-Base Strategy
bf(x)=bg(x)  ⟹  f(x)=g(x)b^{f(x)} = b^{g(x)} \implies f(x) = g(x)

Only works when both sides share the same base

Take-Log-of-Both-Sides Strategy
ax=c  ⟹  x=ln⁡cln⁡aa^x = c \implies x = \frac{\ln c}{\ln a}

Use when bases cannot be matched

Log-to-Exponential Conversion
log⁡b(x)=y  ⟺  by=x\log_b(x) = y \iff b^y = x

The key move for solving logarithmic equations

One-to-One Property of Logarithms
log⁡b(M)=log⁡b(N)  ⟹  M=N\log_b(M) = \log_b(N) \implies M = N

If two logs with the same base are equal, their arguments are equal

Extraneous Solution Check
Domain requirement: arguments of all logs must be >0\text{Domain requirement: arguments of all logs must be } > 0

Always verify solutions in the original equation

Key Takeaways

  • ✓For exponential equations: try to match bases first (bf(x)=bg(x)  ⟹  f(x)=g(x)b^{f(x)} = b^{g(x)} \implies f(x) = g(x)); if you can't, take ln⁡\ln of both sides
  • ✓For log equations: condense to a single log, convert to exponential form, then solve
  • ✓Always check for extraneous solutions in log equations — reject any answer that makes a log argument ≤0\leq 0
  • ✓When you see e2xe^{2x}, think substitution: let u=exu = e^x and solve the resulting quadratic

⚠️ Common Mistakes

  • ✗Forgetting to check for extraneous solutions in log equations — always plug answers back in and reject any that make a log argument ≤0\leq 0
  • ✗Trying to take ln⁡\ln of both sides when the bases can be matched — check for same-base first, it's faster and avoids decimals
  • ✗Distributing a log across addition: log⁡(x+3)≠log⁡(x)+log⁡(3)\log(x + 3) \neq \log(x) + \log(3) — there is no rule for the log of a sum
  • ✗Forgetting to condense before converting: if you have two logs on one side, combine them into one log first, then convert to exponential form