Example

$2^{3x} = 2^{15}$ $3x = 15$ $x = 5$

Strategy 2: Taking Logarithms

If you can't easily get the same base, take the logarithm of both sides.

Steps:

1. Isolate the exponential expression
2. Take log (or ln) of both sides
3. Use the power rule: $\log(b^x) = x\log(b)$
4. Solve for the variable

Example

$5^x = 17$ $\ln(5^x) = \ln(17)$ $x\ln(5) = \ln(17)$ $x = \frac{\ln(17)}{\ln(5)}$

Solving Logarithmic Equations

A logarithmic equation contains logarithmic expressions.

Strategy 1: Convert to Exponential Form

Use the definition: $\log_b(x) = y$ means $b^y = x$

Example

$\log_3(x) = 4$ $x = 3^4 = 81$

Strategy 2: Combine Logarithms

Use logarithm properties to combine into a single log, then solve.

Key Properties to Use

• $\log_b(M) + \log_b(N) = \log_b(MN)$
• $\log_b(M) - \log_b(N) = \log_b(M/N)$
• $p\log_b(M) = \log_b(M^p)$

Strategy 3: Equal Logs Method

If $\log_b(M) = \log_b(N)$, then $M = N$

(assuming same base and same domain)

Important Reminders

⚠️ Check your answers!

• Logarithms require positive arguments: $x > 0$ for $\log(x)$
• Reject any solutions that give $\log(\text{negative})$ or $\log(0)$

⚠️ One-to-one property

• $b^x$ is one-to-one (equal outputs → equal inputs)
• $\log_b(x)$ is one-to-one (equal outputs → equal inputs)

Common Equations to Recognize

Type 1: $b^x = a$

$x = \log_b(a) = \frac{\ln(a)}{\ln(b)}$

Type 2: $a \cdot b^{cx} + d = e$

1. Isolate: $b^{cx} = \frac{e - d}{a}$
2. Take log: $cx\ln(b) = \ln\left(\frac{e - d}{a}\right)$
3. Solve: $x = \frac{1}{c\ln(b)}\ln\left(\frac{e - d}{a}\right)$

Type 3: $\log(x) + \log(x - 3) = 1$

1. Combine: $\log(x(x - 3)) = 1$
2. Convert: $x(x - 3) = 10^1$
3. Solve quadratic: $x^2 - 3x - 10 = 0$

Applications

1. Compound Interest: $A = P(1 + r)^t$
2. Exponential Growth/Decay: $A = A_0e^{kt}$
3. Half-life Problems: $A = A_0(1/2)^{t/h}$
4. Doubling Time: Solve $2A_0 = A_0e^{kt}$ for $$

$5^{2x-1} = 5^3$

Since the bases are equal: $2x - 1 = 3$

$2x = 4$

$x = 2$

Part (b): $3e^{4x} = 24$

$e^{4x} = 8$

Take natural log of both sides: $\ln(e^{4x}) = \ln 8$

$4x = \ln 8$

$x = \frac{\ln 8}{4} = \frac{2.079}{4} \approx 0.520$

Part (c): $\log_2(x+3) + \log_2(x-3) = 4$

Use product rule: $\log_2[(x+3)(x-3)] = 4$

$\log_2(x^2 - 9) = 4$

Convert to exponential form: $x^2 - 9 = 2^4 = 16$

$x^2 = 25$

$x = \pm 5$

Check domain: We need $x + 3 > 0$ and $x - 3 > 0$, so $x > 3$.

Therefore: $x = 5$ (reject $x = -5$)

Step 2: Convert to exponential form. $x(x - 3) = 2^2 = 4$

Step 3: Expand and rearrange. $x^2 - 3x = 4$ $x^2 - 3x - 4 = 0$

Step 4: Factor. $(x - 4)(x + 1) = 0$ $x = 4 \text{ or } x = -1$

Step 5: Check both solutions in the original equation.

For $x = 4$: $\log_2(4) + \log_2(4 - 3) = \log_2(4) + \log_2(1)$ $= 2 + 0 = 2$ ✓

For $x = -1$: $\log_2(-1) + \log_2(-4)$ This is undefined (cannot take log of negative numbers) ✗

Answer: $x = 4$ (reject $x = -1$)