solve 5^2x=8^3x-1 for x. round to nearest two decimal places.

1 answer

To solve the equation \(5^{2x} = 8^{3x-1}\) for \(x\), we can use logarithms. Here are the steps:

1. Apply the natural logarithm (ln) to both sides of the equation:

\[
\ln(5^{2x}) = \ln(8^{3x-1})
\]

2. Use the logarithm power rule, \(\ln(a^b) = b \ln(a)\):

\[
2x \ln(5) = (3x - 1) \ln(8)
\]

3. Distribute \(\ln(8)\) on the right-hand side:

\[
2x \ln(5) = 3x \ln(8) - \ln(8)
\]

4. Rearrange the equation to isolate the terms involving \(x\) on one side:

\[
2x \ln(5) - 3x \ln(8) = -\ln(8)
\]

5. Factor out \(x\) from the left-hand side:

\[
x (2 \ln(5) - 3 \ln(8)) = -\ln(8)
\]

6. Solve for \(x\):

\[
x = \frac{-\ln(8)}{2 \ln(5) - 3 \ln(8)}
\]

Now we need to compute the numerical values of the logarithms:

\[
\ln(5) \approx 1.60944
\]

\[
\ln(8) \approx 2.07944
\]

Substitute these values back into the equation:

\[
x = \frac{-2.07944}{2 \times 1.60944 - 3 \times 2.07944}
\]

\[
x = \frac{-2.07944}{3.21888 - 6.23832}
\]

\[
x = \frac{-2.07944}{-3.01944}
\]

\[
x \approx 0.69
\]

Therefore, the solution rounded to the nearest two decimal places is:

\[
x \approx 0.69
\]