Question

How do you solve `lnx=7.25`?

Answer 1

`x=e^7.25~~1408.105`

Explanation:

Remember from basic definitions
`color(white)("XXX")ln(a)=c` means `e^c=a`

Therefore if
`color(white)("XXX")ln(x)=7.25`
then
`color(white)("XXX")x=e^7.25`
(using a calculator we can find the approximation `e^7.25=1408.105`)

Answer 2

`e^7.25`, or `10^7.25`, or `b^7.25`, `b` being the base of the logarithm.

Explanation:

You can only use the fact that the logarithm and the exponential are one the inverse function of the other. This means that

`e^lnx=x` and `ln(e^x)=x`

Using this property, you can put both left and right member at the exponent:

`ln(x)=7.25 \iff e^ln(x)=e^7.25`

But we have just observed that `e^ln(x)=x`, so we have

`x=e^7.25`.

This, of course, assuming that by "log" you meant the natural one. If, for example, you use base 10 logarithm, you should change `e` with `10`, like this:

`log(x)=7.25 \iff 10^log(x)=10^7.25 \iff x=10^7.25`