Solving Exponential Equations Math Example 3

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Example 3

medium
Solve 2x+3=7x−12^{x+3} = 7^{x-1}.

Solution

  1. 1
    Take ln⁡\ln of both sides: (x+3)ln⁡2=(x−1)ln⁡7(x+3)\ln 2 = (x-1)\ln 7. Expand: xln⁡2+3ln⁡2=xln⁡7−ln⁡7x\ln 2 + 3\ln 2 = x\ln 7 - \ln 7.
  2. 2
    Collect xx terms: x(ln⁡2−ln⁡7)=−ln⁡7−3ln⁡2x(\ln 2 - \ln 7) = -\ln 7 - 3\ln 2, so x=−ln⁡7−3ln⁡2ln⁡2−ln⁡7=ln⁡7+3ln⁡2ln⁡7−ln⁡2=ln⁡56ln⁡(7/2)≈3.228x = \frac{-\ln 7 - 3\ln 2}{\ln 2 - \ln 7} = \frac{\ln 7 + 3\ln 2}{\ln 7 - \ln 2} = \frac{\ln 56}{\ln(7/2)} \approx 3.228.

Answer

x=ln⁡7+3ln⁡2ln⁡7−ln⁡2≈3.228x = \frac{\ln 7 + 3\ln 2}{\ln 7 - \ln 2} \approx 3.228
When the variable appears in exponents on both sides with different bases, take logarithms and distribute to isolate xx. This is a standard technique that converts the problem into solving a linear equation.

About Solving Exponential Equations

Solving exponential equations means finding the unknown variable trapped in an exponent by applying logarithms to both sides, using the power rule to bring the exponent down, and then isolating the variable with standard algebra.

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