# What is solution of equation `log_x 2+log_sqrt x 2 = 9` ?

### 2 Answers | Add Yours

Any logarithm `log_a b` can be written as an expression with two logarithm with the same base as `log_a b = (log_m b)/(log_m a)`

`log_x 2 + log_(sqrt x) 2 = 9`

Use m = 2 and write the given logarithms as shown earlier

=> `(log_2 2)/(log_2 x) + (log_2 2)/(log_2 sqrt x) = 9`

=> `1/(log_2 x) + 1/(log_2 x^(1/2)) = 9`

=> `1/(log_2 x) + 1/((1/2)*log_2 x)) = 9`

Let `log_2 x = y`

=> `1/y + 2/y = 9`

=> `3/y = 9`

=> `y = 1/3`

`log_2 x = (1/3)`

=> `x = 2^(1/3)`

**The required value of **`x = 2^(1/3)`

You need to solve for` x` the given equation, hence, you should convert the base of logarithm, using the following formula, such that:

`log_a b =1/(log_b a)`

Reasoning by analogy, yields:

`log_x 2 = 1/(log_2 x)`

`log_(sqrt x) 2 = 1/(log_2 sqrt x) => log_(sqrt x) 2 = 1/(log_2 (x^(1/2))) => log_(sqrt x) 2 = 2/(log_2 x)`

You should rewrite the equation, such that:

`1/(log_2 x) + 2/(log_2 x) = 9`

You need to come up with the following substitution, such that:

`log_2 x = y`

Changing the variable, yields:

`1/y + 2/y = 9 => (1+2)/y = 9 => 3 = 9y => y = 3/9 => y = 1/3`

You need to solve for `x` the equation `log_2 x = y` , such that:

`log_2 x = 1/3 => x = 2^(1/3) => x = root(3) 2`

**Hence, evaluating the solution to the given equation, yields **`x = root(3) 2.`