# Solve : ` log_2(10 - 2^x) = x + 2`

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### 2 Answers

You may consider the right side multiplied by 1 such that:

`log_2 (10-2^x)=(x+2)*1`

You should write `1 = log_2 2` such that:

`log_2 (10-2^x)=(x+2)*log_2 2`

Using the properties of logarithmic function yields:

`log_2 (10-2^x)=log_2 (2^(x+2))`

Equating arguments of logarithms yields:

`10-2^x = 4*2^x`

You need to bring all terms containing `2^x` to the left side,clearing the right side such that:

`10 - 2^x - 4*2^x = 0`

You need to keep only the terms containing 2^x to the left side such that:

`-5*2^x = -10 =gt 2^x = 2`

Equating the exponents yields: x = 1.

**Hence, the solution to the equation is x=1.**

The equation `log_2(10 - 2^x) = x + 2` has to be solved

`log_2(10 - 2^x) = x + 2`

=> `10 - 2^x = 2^(x + 2)`

=> `10 - 2^x = 4*2^x`

=> `5*2^x = 10`

=> `2^x = 10/5 = 2`

=> `x = 1`

**The solution of the equation is x = ` `1**