# Raising reciprocal. Solve for x. `log_x(125/27)=-3/2` , the result is `x=9/25` Why? `(2x)^2/3 = 3` , the result is x=1.04 Why?I dont know why is x=9/25 or x=1.04

lemjay | High School Teacher | (Level 2) Senior Educator

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Note that logrithim functions can be express in exponential form.

Logarithm Function                     Equivalent Exponential function

` log_b a = m `     =======>>             `b^m = a`

Then, we use this to solve the above problem.

#1 . `log_x 125/27 = -3/2`

`x^(-3/2) = 125/27`

Express 125 and 27 with its prime factors ( `125 = 5*5*5 = 5^3`  and `27 = 3*3*3 = 3^3` ).

`x^(-3/2) = 5^3/3^3`

`x^(-3/2) = 5^3/3^3`

Raise both sides by -2/3 to isolate x.

`(x^(-3/2))^(-2/3) = (5^3/3^3)^(-2/3)`

Simplify using the properties of exponents which are `(z^m)^n = z^(m*n)`  and   `(z/y)^m = z^m/y^m` .

`x^((-3/2)(-2/3)) = 5^(3(-2/3))/3^(3(-2/3))`

` x = 5^-2/3^-2 = (1/5^2)/(1/3^2) = 3^2 / 5^2`

Hence,   x = `9/25` .

#2.  `(2x)^2/3 = 3`

`(2x)^2 = 3^2`

Raise both sides by 1/2.

`[(2x)^2]^(1/2) = (3^2)^(1/2)`

Simplify using the property of exponent which is `(z^m)^n = z^(m*n)` .

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

`2x = 3`

`x = 3/2`

Hence, x `= 3/2 = 1.5` .  ` `

Note that for #2, there may be an error either with the given equation or with the given answer. Please verify.