# Solve this exponential equation: 2^(x+1) = e^x

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`2^(x+1) = e^x`

In logarithm we know that` log(x^a) = alogx` . Here instead of `log` we take `ln ` because we have the exponential function here. log and `ln` both satisfy logarithm rules.

`2^(x+1) = e^x`

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

`(x+1)ln2 = xlne`

We know that `log10 = 1` . It is because log is based on base 10. For `ln` it is based on base e. So `lne = 1`

`(x+1)ln2 = xlne`

`(x+1)ln2 = x`

`(x+1)ln2-x = 0`

`xln2+ln2-x = 0`

`x(ln2-1)+ln2 = 0`

`x = -ln2/(ln2-1)`

*So the answer is;*

`x = (-ln2)/(ln2-1)`

The equation 2^(x+1) = e^x has to be solved.

2^(x+1) = e^x

take the logarithm of both the sides and use the property log a^b = b*log a

log(2^(x+1)) = log(e^x)

=> (x + 1)*log 2 = x*log e

=> log 2 = x*(log e - log 2)

=> x = (log 2)/(log e - log 2)

**The solution of the equation is x = (log 2)/(log e - log 2)**