# How do you solve this expression?   `log_x (125/27) = 3/4`

You need to solve the following equation such that:

`log_x (125/27) = 3/4`

You need to solve the equation for x using the logarithmic identity such that:

`log_x a = b => x^b = a`

Reasoning by analogy yields:

`log_x (125/27) = 3/4 => 125/27 = x^(3/4)`

Converting the rational power...

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You need to solve the following equation such that:

`log_x (125/27) = 3/4`

You need to solve the equation for x using the logarithmic identity such that:

`log_x a = b => x^b = a`

Reasoning by analogy yields:

`log_x (125/27) = 3/4 => 125/27 = x^(3/4)`

Converting the rational power into a radical yields:

`x^(3/4) = root(4)(x^3)`

`125/27 = root(4)(x^3)`

You need to remove the radical, hence, you need to raise to the 4th power both sides such that:

`(125/27)^4 = x^3 `

You need to take cube root both sides such that:

`root(3)((125/27)^4) = root(3)(x^3)`

`125/27*root(3)(125/27) = x`

You should write `125 = 5^3`  and `27 = 3^3`  such that:

`125/27*root(3)(5^3/3^3) = x`

Converting the cube root into a rational power yields:

`125/27*((5/3)^3)^(1/3) = x`

`x = 125/27*(5/3) => x = 625/81`

Since `625/81>0`  and `625/81!=1` , hence, evaluating the solutions to the given equation yields `x = 625/81` .

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