How do solve equation `25^(log_5 (lg x))=lgx-lg^2x+1?`

### 1 Answer | Add Yours

You should use the following logarithmic identity `n*log x = log x^n` to solve `25^(log_5 (lg x))` such that:

`25^(log_5 (lg x)) = 5^(2log_5 (lg x))`

You need to use the logarithmic identity `a^(log_a b) = b` such that:

`5^(2log_5 (lg x)) = 5^(log_5 (lg x)^2) = lg^2 x`

Substituting `lg^2 x` for `5^(2log_5 (lg x))` yields:

`lg^2 x = lg x - lg^2 x + 1 => 2lg^2 x - lg x - 1 = 0`

You should come up with the following substitution `lg x = y` such that:

`2y^2 - y - 1 = 0`

Using quadratic formula yields:

`y_(1,2) = (1+-sqrt(1+8))/4 => y_(1,2) = (1+-sqrt9)/4`

`y_(1,2) = (1+-3)/4 => y_1 = 1 ;y_2 = -1/2`

You need to solve for x the following equations `lg x = y_(1,2)` such that:

`lg x = 1 => x = 10^1 => x = 10 `

`lg x = -1/2 => x = 10^(-1/2) => x = 1/(10^(1/2)) => x = 1/sqrt10`

**Hence, evaluating the solutions to the given equation yields `x = 1/sqrt10` and `x = 10` .**

**Sources:**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes