Question

log[3](x-2)=log[3] 27 - log[3](x-4) -5^(log[5]1)
I have tried this out and right now all I am getting is
log[3] (x-2)(x-4)/27= -5^(log[5] 1)
Is this somewhat right so far? I'm not sure what to do with the "log[3] 27"
the answer is supposed to be- 3+sqrt(10)

Answer 1

hint :
-5^(log[5] 1 = -1
now, setting -1 be log[3] (1/3)

Answer 2

HINT(also): \(\large log_327=a \Leftrightarrow 3^a=27 \)
what's a= ???

Answer 3

a=3? D:

Answer 4

yes, that's right.. then, ?

Answer 5

well, i want make easier for u
from ur solution above :
log[3] (x-2)(x-4)/27= -5^(log[5] 1
setting right side be -1
it can be :
log[3] (x-2)(x-4)/27= -1 then
-1 = [3] (1/3), therefore it can be
log[3] (x-2)(x-4)/27= [3] (1/3)
cancel out the log[3] to both sides
(x-2)(x-4)/27= 1/3 or
(x-2)(x-4) = 27 * 1/3
(x-2)(x-4) = 9
now, solve for x

Answer 6

* log[3] (x-2)(x-4)/27= log [3] (1/3)

Answer 7

Ok! I was close. I for for to put log[3] on both sides. But now that I see where I went wrong, I will finish solving for x and let you know if I have any problems. Thank you so much for your help!

Answer 8

*forgot to put log[3]

Answer 9

okay...
to get solution of (x-2)(x-4) = 9
simplify, frist. then use quadratic formuka :
x = {-b +- sqrt(b^2-4ac)}/2a
if u got 2 solution's, that one positive and the other negative, just take that positive value, because for x negative undefined for log[3] (x-2)

Answer 10

I finished the problem! I got the correct answer! Thanks for your help! One last question. for -5^(log[5] 1, did you plug that in a calculator? I tried that and I'm getting the number 3.

Answer 11

no, i have used the property of logarithm :