Solving Exponential Equations Math Example 2

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Example 2

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Solve 52xโˆ’1=125^{2x-1} = 12.

Solution

  1. 1
    Take the natural logarithm of both sides: lnโก(52xโˆ’1)=lnโก(12)\ln(5^{2x-1}) = \ln(12).
  2. 2
    Apply the power rule: (2xโˆ’1)lnโก(5)=lnโก(12)(2x-1)\ln(5) = \ln(12).
  3. 3
    Solve for xx: 2xโˆ’1=lnโก(12)lnโก(5)2x - 1 = \frac{\ln(12)}{\ln(5)}, so 2x=lnโก(12)lnโก(5)+12x = \frac{\ln(12)}{\ln(5)} + 1.
  4. 4
    x=12(lnโก12lnโก5+1)=12(2.48491.6094+1)โ‰ˆ12(1.5440+1)โ‰ˆ1.272x = \frac{1}{2}\left(\frac{\ln 12}{\ln 5} + 1\right) = \frac{1}{2}\left(\frac{2.4849}{1.6094} + 1\right) \approx \frac{1}{2}(1.5440 + 1) \approx 1.272.

Answer

x=lnโก12+lnโก52lnโก5=lnโก602lnโก5โ‰ˆ1.272x = \frac{\ln 12 + \ln 5}{2\ln 5} = \frac{\ln 60}{2\ln 5} \approx 1.272
When the two sides cannot be written with the same base, take logarithms of both sides. The power rule brings the variable exponent down, converting the exponential equation into a linear equation in xx.

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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