Solve `2^(x+1)=5` :

Take a logarithm of both sides. You can use any base -- typically you would choose to use base `e` (i.e. the natural logarithm) or base 10 (the common logarithm) as most scientific calculators can compute these.

Using the natural logarithm `ln` we get:

`ln(2^(x+1))=ln5`

A property of logarithms, of any base,is `log a^b=bloga` . So:

`(x+1)ln2=ln5`

`x+1=ln5/ln2`

`x=ln5/ln2 -1`

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The solution to `2^(x+1)=5` is `x=ln5/ln2-1~~1.322`

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Note that the answer provided by mjripalda is fundamentally the same -- you can use the change of base formula to convert `log_2 5` to `ln5/ln2` , or `log5/log2` , or any other base of your choosing.

Note that an exponential expression has its equivalent logarithm form.

Exponential Expression <====> Logarithmic Expresion

`b^m = a` <====> `log_b a = m`

So,

`2^(x+1) = 5`

It's equivalent logarithm form is,

`log_2 5 = x+1`

Then, we may solve for x.

`log_2 5 - 1= x`

**Answer: **x = `log_2 5` - 1** **