Solve algebraically (without using logarithms):

(1/8)^2x = 16^(x-5)

Question

Solve algebraically (without using logarithms):

(1/8)^2x = 16^(x-5)

anonymous · Answer

The way to solve it without logarithms is to have the same base.

thesmartone · Answer

Hi there! @lisamath12 

Nice to meet you again! :)

thesmartone · Answer

Ok, so we need the same base, because we cannot solve this using logarithms. And the only other way it to get the same base.

Do you agree? :)

thesmartone · Answer

Ok, I'll give you a hint since we don't want you to be stuck at the same spot for forever. :)

$\sf\Large \frac{1}{8} = \frac{1}{2^3} = 2^{-3}$

Second hint:

$\sf\Large 16 = 2^4$

Can you use this information to make both sides of the equation have the same base?

Remember, if $\sf a^b = a^c$ then that means b = c

anonymous · Answer

@lisamath12 Did you get anything that I have said above? :)

thesmartone · Answer

@lisamath12

Hello? No response from you, and it's been 2 days.