Exponential and Logarithmic Equations

📌 Overview

Exponential and logarithmic equations require special techniques to solve because the variable appears in the exponent or inside a logarithm.

Solving Exponential Equations by Rewriting the Base

One-to-One Property of Exponents:

If the bases are equal, the exponents must be equal:

If $ b^x = b^y $, then $ x = y $

Where $ b > 0 $ and $ b \neq 1 $

Steps to Solve:

Rewrite each side as a power with the same base
Use exponent rules to simplify if necessary
Set the exponents equal to each other
Solve the resulting equation
Check your solution in the original equation

📝 Examples - Same Base Method:

Example 1: Solve $ 2^{3x+1} = 8 $

Rewrite 8 as a power of 2: $ 8 = 2^3 $
$ 2^{3x+1} = 2^3 $
Set exponents equal: $ 3x + 1 = 3 $
Solve: $ 3x = 2 $ → $ x = \frac{2}{3} $

Example 2: Solve $ 9^{x-2} = 27^x $

Rewrite with base 3: $ (3^2)^{x-2} = (3^3)^x $
Simplify: $ 3^{2(x-2)} = 3^{3x} $
$ 3^{2x-4} = 3^{3x} $
Set exponents equal: $ 2x - 4 = 3x $
Solve: $ -4 = x $ → $ x = -4 $

Example 3: Solve $ 4^{x+1} = \frac{1}{16} $

Rewrite: $ (2^2)^{x+1} = 2^{-4} $
Simplify: $ 2^{2(x+1)} = 2^{-4} $
$ 2^{2x+2} = 2^{-4} $
Set exponents equal: $ 2x + 2 = -4 $
Solve: $ 2x = -6 $ → $ x = -3 $

Solving Exponential Equations Using Logarithms

When to Use Logarithms:

When bases cannot be easily rewritten to be the same, use logarithms to solve.

Steps:

Isolate the exponential expression
Take the logarithm of both sides (common log or natural log)
Use the power property of logarithms to bring down the exponent
Solve for the variable
Approximate using a calculator if needed

📝 Examples - Using Common Logarithms:

Example 1: Solve $ 5^x = 20 $

Take log of both sides: $ \log(5^x) = \log(20) $
Use power property: $ x \log 5 = \log 20 $
Solve for x: $ x = \frac{\log 20}{\log 5} $
Calculate: $ x = \frac{1.301}{0.699} \approx 1.861 $

Example 2: Solve $ 3 \cdot 2^{x+1} = 48 $

Isolate exponential: $ 2^{x+1} = 16 $
Take log: $ \log(2^{x+1}) = \log 16 $
Use power property: $ (x+1) \log 2 = \log 16 $
Solve: $ x + 1 = \frac{\log 16}{\log 2} = \frac{1.204}{0.301} = 4 $
$ x = 3 $

Example 3: Solve $ 7^{2x-3} = 100 $

Take log: $ \log(7^{2x-3}) = \log 100 $
Power property: $ (2x-3) \log 7 = 2 $
Solve: $ 2x - 3 = \frac{2}{\log 7} = \frac{2}{0.845} \approx 2.367 $
$ 2x = 5.367 $
$ x \approx 2.684 $

📝 Examples - Using Natural Logarithms:

Example 1: Solve $ e^x = 10 $

Take ln of both sides: $ \ln(e^x) = \ln 10 $
Simplify: $ x = \ln 10 \approx 2.303 $

Example 2: Solve $ 2e^{3x} = 50 $

Isolate: $ e^{3x} = 25 $
Take ln: $ \ln(e^{3x}) = \ln 25 $
Simplify: $ 3x = \ln 25 $
Solve: $ x = \frac{\ln 25}{3} = \frac{3.219}{3} \approx 1.073 $

Example 3: Solve $ e^{2x-1} = 7 $

Take ln: $ \ln(e^{2x-1}) = \ln 7 $
Simplify: $ 2x - 1 = \ln 7 $
Solve: $ 2x = \ln 7 + 1 = 1.946 + 1 = 2.946 $
$ x \approx 1.473 $

Compound Interest

Compound Interest Formula:

$ A = P\left(1 + \frac{r}{n}\right)^{nt} $

Where:

$ A $ = Final amount (future value)
$ P $ = Principal (initial amount)
$ r $ = Annual interest rate (as a decimal)
$ n $ = Number of times interest is compounded per year
$ t $ = Time in years

Common Compounding Frequencies:

Annually: $ n = 1 $
Semi-annually: $ n = 2 $
Quarterly: $ n = 4 $
Monthly: $ n = 12 $
Weekly: $ n = 52 $
Daily: $ n = 365 $

📝 Example - Compound Interest:

You deposit $2,000 in an account paying 5% annual interest compounded quarterly. How much will you have after 3 years?

Given:

$ P = 2000 $, $ r = 0.05 $, $ n = 4 $, $ t = 3 $

Solution:

$ A = 2000\left(1 + \frac{0.05}{4}\right)^{4 \cdot 3} $
$ = 2000(1.0125)^{12} $
$ = 2000(1.1608) $
$ = \$2,321.60 $

Interest earned: $ \$2,321.60 - \$2,000 = \$321.60 $

Continuously Compounded Interest

Continuous Compounding Formula:

When interest is compounded continuously (infinite number of times):

$ A = Pe^{rt} $

Where:

$ A $ = Final amount
$ P $ = Principal
$ e $ = Euler's number (≈ 2.71828)
$ r $ = Annual interest rate (as a decimal)
$ t $ = Time in years

📝 Example - Continuous Compounding:

Invest $3,500 at 4.5% annual interest compounded continuously for 6 years. Find the final amount.

Given:

$ P = 3500 $, $ r = 0.045 $, $ t = 6 $

Solution:

$ A = 3500e^{0.045 \cdot 6} $
$ = 3500e^{0.27} $
$ = 3500(1.3100) $
$ = \$4,585.00 $

📝 Example - Finding Time:

How long will it take $1,000 to double at 6% interest compounded continuously?

Given: $ P = 1000 $, $ A = 2000 $, $ r = 0.06 $
Formula: $ 2000 = 1000e^{0.06t} $
Divide by 1000: $ 2 = e^{0.06t} $
Take ln: $ \ln 2 = 0.06t $
Solve: $ t = \frac{\ln 2}{0.06} = \frac{0.693}{0.06} \approx 11.55 $ years

Solving Logarithmic Equations

Two Main Methods:

Method 1: Convert to Exponential Form

If $ \log_b x = y $, then $ b^y = x $

Method 2: Use Properties of Logarithms

Combine logs using properties, then solve

General Steps:

Isolate the logarithmic expression(s)
Use properties to combine into a single log if possible
Convert to exponential form OR use one-to-one property
Solve the resulting equation
Check for extraneous solutions (arguments must be positive)

📝 Example 1 - Single Log:

Solve $ \log_3(2x - 1) = 4 $

Convert to exponential form: $ 3^4 = 2x - 1 $
Simplify: $ 81 = 2x - 1 $
Solve: $ 82 = 2x $
$ x = 41 $

Check: $ \log_3(2(41) - 1) = \log_3 81 = 4 $ ✓

📝 Example 2 - Using Properties:

Solve $ \log_2 x + \log_2(x - 3) = 2 $

Use product property: $ \log_2[x(x-3)] = 2 $
Convert: $ x(x-3) = 2^2 $
Simplify: $ x^2 - 3x = 4 $
$ x^2 - 3x - 4 = 0 $
Factor: $ (x-4)(x+1) = 0 $
Solutions: $ x = 4 $ or $ x = -1 $

Check:

$ x = 4 $: $ \log_2 4 + \log_2 1 = 2 + 0 = 2 $ ✓
$ x = -1 $: $ \log_2(-1) $ is undefined ✗

Solution: $ x = 4 $ (x = -1 is extraneous)

📝 Example 3 - Logs on Both Sides:

Solve $ \log(3x + 1) = \log(2x + 5) $

Use one-to-one property: $ 3x + 1 = 2x + 5 $
Solve: $ x = 4 $

Check: $ \log 13 = \log 13 $ ✓

📝 Example 4 - Natural Logarithm:

Solve $ \ln(x + 2) - \ln x = 1 $

Use quotient property: $ \ln\left(\frac{x+2}{x}\right) = 1 $
Convert: $ \frac{x+2}{x} = e^1 $
Simplify: $ \frac{x+2}{x} = e $
Multiply by x: $ x + 2 = ex $
Solve: $ 2 = ex - x = x(e-1) $
$ x = \frac{2}{e-1} = \frac{2}{1.718} \approx 1.164 $

📝 Example 5 - Quadratic Form:

Solve $ (\log x)^2 = \log x^2 $

Use power property: $ (\log x)^2 = 2\log x $
Rearrange: $ (\log x)^2 - 2\log x = 0 $
Factor: $ \log x(\log x - 2) = 0 $
Solutions: $ \log x = 0 $ or $ \log x = 2 $

If $ \log x = 0 $: $ x = 10^0 = 1 $
If $ \log x = 2 $: $ x = 10^2 = 100 $

Solutions: $ x = 1 $ or $ x = 100 $

Extraneous Solutions

Why Check for Extraneous Solutions?

When solving logarithmic equations:

Arguments must be positive: Cannot take log of zero or negative numbers
Always check solutions in the original equation
Discard any solution that makes an argument ≤ 0
Algebraic manipulation can introduce extraneous solutions

⚡ Quick Summary

Exponential equations (same base): Rewrite with common base, set exponents equal
Exponential equations (different bases): Take log of both sides, use power property
Compound interest: $ A = P(1 + \frac{r}{n})^{nt} $
Continuous compounding: $ A = Pe^{rt} $
Logarithmic equations: Use properties to combine, convert to exponential form
Always check for extraneous solutions in log equations
Log arguments must always be positive

📚 Key Formulas Reference

Exponential Equations: