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

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lemjay | High School Teacher | (Level 3) 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.

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