Solve the exponential equation. 9^8x = 27 I don't understand how to solve this. If you know I will give medal (: A) 5/8 B) 3/16 C) 9 D) 16/3
hint take logarithns
Hint: rewrite both the 9 and the 27 as powers of 3:\[(3^{2})^{8x}=3^{3}\] since y ou now have the same base on both sides, it's easy to solve for x. Try it. If the resulting x doesn't match any of your answers, double check to ascertain that you've copied the problem down correctly.
so the logarithm would look like this right? log8^27 = x
or you can convert as follows - i was just going to do what mathmale has just written
alright. I will solve and then tell you what I got. Then you or mathmale will tell me if I'm right.
no 8x *log 9 = log 27
If you, usernae, choose to use logs to solve this problem for x, you'd end up with 8x log 9 = log 27. Solve for x.
so do I divide log 9 from log 27 in order to be left with x. Then simplify from there?
Yes, that's an appropriate approach. Then you'll need to isolate x.
my answer is x = 3/8
Please check that result by substitution.
no thats not the answer show us your working
I divided log27/log9
x should be = [log 27 / log 9] / 8
then I isolated the x by dividing both sides by 8
right so far
yep
but that's not one of the answer choices. However, when I plugged in the logarithm into wolfram it gave me 3/16. Where did the 16 come from? Do you know?
what did you get log 27 / log 9?
yes I did (:
If \[9^{8x}=27,\] \[(3^{2})^{8x}=3^{3}\] is also true. Then 16x=3, and x = ???
Logs should also lead to the correct answer.
log27/lgo9 = 3/2 divide this by 8 gives 3/2 * 1/8 = 3/16
I advocated this particular solution some time ago, but read it incorrectly and tried to solve 6x=3 for x, whereas it should be 16x=3.
but wait. 27/9 = 3.
While 27/9=3, you are actually dividing (log 27) by (log 9).
you need to divide the logs not the numbers log 27 / log 9 = 1.5 or 3/2
And that 's not 3.
ohhh okay then. I get it now. so the answer is 3/16. Thank you @cwrw238! and thank you mathmale! Listen if it's not too much, I have one more question.
Post it separately, please!
Join our real-time social learning platform and learn together with your friends!