how to solve for questions like 5-e^2x=-15

$$5 - e^{2x}=-15$$

$$20=e^{2x}$$

$$\ln(20)=2x$$

$$x=\dfrac {\ln(20)} 2$$

$${\frac{{ln}{\left({\mathtt{20}}\right)}}{{\mathtt{2}}}} = {\mathtt{1.497\: \!866\: \!136\: \!776\: \!995\: \!5}}$$

how to solve for questions like 5-e^2x=-15

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I'm going to assume that you're asking this:

5 - e^(2x) = -15         .......subtract 5 from both sides and then multiply both sides by (-1)

-e^2x = -20

 e^(2x) = 20                 Now, take the "natural log," (ln), of both sides

ln e^(2x) =  ln 20          By a property of logs, the "2x" can be "brought out front" on the left side

(2x) lne = ln 20          and "lne" just equals 1     so we have

2x = ln 20                  so.......

x = ln 20 / 2  ≈  1.5