Solve (40!)=5^2x

$$(40 ! )= { 5 }^{ 2x }$$

Solve for x

$$x = \frac{\log_{5} {(815915283247897734345611269596115894272000000000)}}{2} \approx 34.273033729$$

Steps Using Definition of Logarithm

The factorial of 40 is 815915283247897734345611269596115894272000000000.

$$815915283247897734345611269596115894272000000000=5^{2x}$$

Swap sides so that all variable terms are on the left hand side.

$$5^{2x}=815915283247897734345611269596115894272000000000$$

Take the logarithm of both sides of the equation.

$$\log(5^{2x})=\log(815915283247897734345611269596115894272000000000)$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$2x\log(5)=\log(815915283247897734345611269596115894272000000000)$$

Divide both sides by \log(5).

$$2x=\frac{\log(815915283247897734345611269596115894272000000000)}{\log(5)}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$2x=\log_{5}\left(815915283247897734345611269596115894272000000000\right)$$

Divide both sides by 2.

$$x=\frac{\log_{5}\left(815915283247897734345611269596115894272000000000\right)}{2}$$

Solve for x (complex solution)

$$x=\frac{\pi n_{1}i}{\ln(5)}+\frac{\log_{5}\left(417748625022923639984952970033211337867264\right)}{2}+\frac{9}{2}$$

$$n_{1}\in \mathrm{Z}$$