Question

I need help on solving exponential equations, because I wasn't in class yesterday while we learned it. Please help me!?
25^2x+3=25^5x-9

Answer 1

The equation is this right?
\[\Large 25^{2x+3} = 25^{5x-9}\]

Answer 2

If so, then we can equate the exponents. This is because the bases are equal, so in order for both sides to be equal, the exponents must be equal

that means

\[\Large 25^{2x+3} = 25^{5x-9}\]
leads to

\[\Large 2x+3 = 5x-9\]

solve for x to get your answer

Answer 3

this is assuming my initial assumption about the problem is correct

Answer 4

Yes that is the right equation I typed it on my phone so it was weird, and thank you so much! That helped a lot!

Answer 5

I'm glad it makes more sense now

Answer 6

Okay so now can you please help me on how to get the bases equal? please?
\[9^{8x-4}=81^{3x+6}\]

Answer 7

the bases aren't equal, so we can't equate the exponents like we did in the prev problem

but there's a simple fix here

notice how 81 is equal to 9*9 or 9^2

so we can do this bit of algebraic manipulation

\[\Large 9^{8x-4} = 81^{3x+6}\]

\[\Large 9^{8x-4} = \color{red}{81}^{3x+6}\]

\[\Large 9^{8x-4} = (\color{red}{81})^{3x+6}\]

\[\Large 9^{8x-4} = \left(\color{red}{9^2}\right)^{3x+6}\]

\[\Large 9^{8x-4} = \left(9^2\right)^{3x+6}\]

\[\Large 9^{8x-4} = 9^{2(3x+6)}\]

Answer 8

On that last step, I used the rule \[\Large \left(a^b\right)^c = a^{b*c}\]

Answer 9

now that the bases are both equal to 9, you can now equate the exponents and solve for x

Answer 10

Thank you so much. Definitely understand it better now.

Answer 11

you're welcome

Answer 12

Can you help me work through solving a logarithm equation? Please.
\[1/2\log6(25)+\log6{x} =\log6{20} \]

Answer 13

the '6' is the base of each log?