`5^(6x) = 8320` Solve the equation accurate to three decimal places

For exponential equation: `5^(6x)= 8320` , we may apply the logarithm property:

`log(x^y) = y * log (x).`

This helps to bring down the exponent value.

 Taking "log" on both sides:

`log(5^(6x))=log(8320)`

`6x * log (5) = log(8320)`

Divide both sides by log (5) to isolate "6x":

`(6x * log (5)) /(log(5))= (log(8320))/(log(5))`

`6x=(log(8320))/(log(5))`

Multiply both sides by `1/6` to isolate x:

`(1/6)*6x=(log(8320))/(log(5))*(1/6)`

`x =(log(8320))/(6log(5))`

`x=0.935`