Here's the question you clicked on:
Jadelynn
Solve the exponential equation 9^2x = 27
9 = 3^2 9^2x = 3^4x 27 = 3^3 Compare 4x = 3 x = 3/4
We have\[9^{2x}=27\] \[(3^{2})^{2x}=3^{3}\] \[3^{4x} = 3^{3}\] so 4x = 3
Can I get an explanation on how you did that?
9 is nothing but 3^2 and 27 is 3^3 so we need to make the base common to make it possible to solve for 'x'.... so............... 3^2(2x) = 3^3 therefore 3^4x = 3^3 since the base is the same, the exponents are equal;( since it is an equation and the left hand side is equal to the right hand side). so, 4x = 3 hence x = 3/4
did u understand my explanation @Jadelynn ? because if u didnt i can explain again
Good explanation, @ShikhaDessai
thank you so much!!!
how do you solve the equation 9^2x = 27
using exponential and logarithmic form
@66 maybe post a new question... this one is four months old.
not mine please help if posible
"how do you solve the equation 9^2x = 27" Or... just read above... your question is already answered above and explained twice...
If you need to do it with logarithms, then take logarithms of both sides of 9^2x = 27
how do you do that and its not solved above
can you please help if not then i will google search it , but thanks for your help anyway
does anyone know how to solve for the equation 9^2x = 27