Answered By Suman G

 $\log_bX=Y$log_bX=Y  is equivalent to X=b^Y.

According to this formula, in our case, X=125^2/3 = (5³)^2/3 = 5² = 25

Hence, X =25

Answered By Mingrui (Mark) J

log _b Q = (ln b) / (ln Q)

log₁₂₅ X = (ln X) / (ln 125) = 2 / 3

ln X = (2/3) * (ln 125) = ln (125 ^2/3 ) 

125^2/3 = (³√125) ² = 5² = 25

ln X = ln(125 ^2/3) = ln(25)

ln X = ln 25

X = 5

(you can use base-10 log which is just "log" instead of ln and that would work too)

:)

Answered By Mingrui (Mark) J

sorry, last post is x = 25 i meant

Answered By Daiwei L

 Since this logarithem form is difficult to work with, we change it to exponential. 

The exponential is an inverse of the logarithem function. Therefore, we swapthe x and y of (log₁₂₅x = 2/3). 

Thus, we have 125 as the base, 2/3 as the power and we have 125^2/3 = x.

We can then solve for x, 125^2/3 = 5^{3 * (2/3)} = 5² = 25 = x

Answered By Aleksandar B

The form y=b^x can also be written as log_b(y)=x_. 

So log₁₂₅(x)= 2/3 translates to x= 125^2/3=(125²)^1/3=25.

Answered By Jackie C

X=(125)^2/3=[(125)^1/3]²=[(5*5*5)^1/3]²=(5)²=25