Solving Exponential and Logarithmic Equations

Techniques for solving equations involving exponentials and logarithms

Solving Exponential and Logarithmic Equations

Solving Exponential Equations

An exponential equation has the variable in the exponent.

Strategy 1: Same Base Method

If you can express both sides with the same base, set the exponents equal.

If bx=byb^x = b^y, then x=yx = y

Example

23x=2152^{3x} = 2^{15} 3x=153x = 15 x=5x = 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⁡(bx)=xlog⁡(b)\log(b^x) = x\log(b)
  4. Solve for the variable

Example

5x=175^x = 17 ln⁡(5x)=ln⁡(17)\ln(5^x) = \ln(17) xln⁡(5)=ln⁡(17)x\ln(5) = \ln(17) x=ln⁡(17)ln⁡(5)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\log_b(x) = y means by=xb^y = x

Example

log⁡3(x)=4\log_3(x) = 4 x=34=81x = 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(MN)
  • log⁡b(M)−log⁡b(N)=log⁡b(M/N)\log_b(M) - \log_b(N) = \log_b(M/N)
  • plog⁡b(M)=log⁡b(Mp)p\log_b(M) = \log_b(M^p)

Strategy 3: Equal Logs Method

If log⁡b(M)=log⁡b(N)\log_b(M) = \log_b(N), then M=NM = N

(assuming same base and same domain)

Important Reminders

⚠️ Check your answers!

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

⚠️ One-to-one property

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

Common Equations to Recognize

Type 1: bx=ab^x = a

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

Type 2: a⋅bcx+d=ea \cdot b^{cx} + d = e

  1. Isolate: bcx=e−dab^{cx} = \frac{e - d}{a}
  2. Take log: cxln⁡(b)=ln⁡(e−da)cx\ln(b) = \ln\left(\frac{e - d}{a}\right)
  3. Solve: x=1cln⁡(b)ln⁡(e−da)x = \frac{1}{c\ln(b)}\ln\left(\frac{e - d}{a}\right)

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

  1. Combine: log⁡(x(x−3))=1\log(x(x - 3)) = 1
  2. Convert: x(x−3)=101x(x - 3) = 10^1
  3. Solve quadratic: x2−3x−10=0x^2 - 3x - 10 = 0

Applications

  1. Compound Interest: A=P(1+r)tA = P(1 + r)^t
  2. Exponential Growth/Decay: A=A0ektA = A_0e^{kt}
  3. Half-life Problems: A=A0(1/2)t/hA = A_0(1/2)^{t/h}
  4. Doubling Time: Solve 2A0=A0ekt2A_0 = A_0e^{kt} for tt

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve for xx: 32x−1=273^{2x-1} = 27

💡 Show Solution

Solution:

Step 1: Express both sides with the same base.

Notice that 27=3327 = 3^3: 32x−1=333^{2x-1} = 3^3

Step 2: Set exponents equal.

Since the bases are equal: 2x−1=32x - 1 = 3

Step 3: Solve for xx. 2x=42x = 4 x=2x = 2

Step 4: Check. 32(2)−1=34−1=33=273^{2(2)-1} = 3^{4-1} = 3^3 = 27 ✓

Answer: x=2x = 2

2Problem 2medium

❓ Question:

Solve the following equations:

a) 52x−1=1255^{2x-1} = 125 b) 3e4x=243e^{4x} = 24 c) log⁡2(x+3)+log⁡2(x−3)=4\log_2(x+3) + \log_2(x-3) = 4

💡 Show Solution

Solution:

Part (a): 52x−1=1255^{2x-1} = 125

Rewrite 125 as a power of 5: 125=53125 = 5^3

52x−1=535^{2x-1} = 5^3

Since the bases are equal: 2x−1=32x - 1 = 3

2x=42x = 4

x=2x = 2

Part (b): 3e4x=243e^{4x} = 24

e4x=8e^{4x} = 8

Take natural log of both sides: ln⁡(e4x)=ln⁡8\ln(e^{4x}) = \ln 8

4x=ln⁡84x = \ln 8

x=ln⁡84=2.0794≈0.520x = \frac{\ln 8}{4} = \frac{2.079}{4} \approx 0.520

Part (c): log⁡2(x+3)+log⁡2(x−3)=4\log_2(x+3) + \log_2(x-3) = 4

Use product rule: log⁡2[(x+3)(x−3)]=4\log_2[(x+3)(x-3)] = 4

log⁡2(x2−9)=4\log_2(x^2 - 9) = 4

Convert to exponential form: x2−9=24=16x^2 - 9 = 2^4 = 16

x2=25x^2 = 25

x=±5x = \pm 5

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

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

3Problem 3medium

❓ Question:

Solve for xx: 5x=235^x = 23

💡 Show Solution

Solution:

Step 1: Take the natural log of both sides.

ln⁡(5x)=ln⁡(23)\ln(5^x) = \ln(23)

Step 2: Use the power rule. xln⁡(5)=ln⁡(23)x\ln(5) = \ln(23)

Step 3: Solve for xx. x=ln⁡(23)ln⁡(5)x = \frac{\ln(23)}{\ln(5)}

Step 4: Calculate (optional). x≈3.1351.609≈1.948x \approx \frac{3.135}{1.609} \approx 1.948

Answer: x=ln⁡(23)ln⁡(5)≈1.948x = \frac{\ln(23)}{\ln(5)} \approx 1.948

4Problem 4medium

❓ Question:

Solve for xx: log⁡2(x)+log⁡2(x−3)=2\log_2(x) + \log_2(x - 3) = 2

💡 Show Solution

Solution:

Step 1: Combine logarithms using the product rule.

log⁡2(x)+log⁡2(x−3)=log⁡2(x(x−3))=2\log_2(x) + \log_2(x - 3) = \log_2(x(x - 3)) = 2

Step 2: Convert to exponential form. x(x−3)=22=4x(x - 3) = 2^2 = 4

Step 3: Expand and rearrange. x2−3x=4x^2 - 3x = 4 x2−3x−4=0x^2 - 3x - 4 = 0

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

Step 5: Check both solutions in the original equation.

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

For x=−1x = -1: log⁡2(−1)+log⁡2(−4)\log_2(-1) + \log_2(-4) This is undefined (cannot take log of negative numbers) ✗

Answer: x=4x = 4 (reject x=−1x = -1)

5Problem 5hard

❓ Question:

Solve: log₂(x + 3) + log₂(x - 3) = 4

💡 Show Solution

Step 1: Use product rule to combine logs: log₂[(x + 3)(x - 3)] = 4

Step 2: Simplify: log₂(x² - 9) = 4

Step 3: Convert to exponential form: x² - 9 = 2⁴ x² - 9 = 16

Step 4: Solve for x: x² = 25 x = ±5

Step 5: Check domain restrictions: For x = 5: x + 3 = 8 > 0 ✓, x - 3 = 2 > 0 ✓ For x = -5: x + 3 = -2 < 0 ✗ x = -5 is extraneous

Step 6: Verify x = 5: log₂(8) + log₂(2) = 3 + 1 = 4 ✓

Answer: x = 5

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