Question

Another logarithm + indices question...

Answer 1

Question?

Answer 2

Sorry for asking multiple questions in the same vein, but this is proving to be a tricky subject for me and I just can't understand it properly by myself. u_u

Solve for real x the equation:

\[4(3^{2x+1}) + 17(3^{x}) - 7 = 0\]

I did a few things, ultimately ending up with the equation:

\[4\log_{3} 3^{3x+1} + 17\log_{3} 3^{x} = \log_{3} 7 \]

I must have got it wrong somewhere because the answer I get from this is nowhere near to my textbook's answer. Any hints?

Thanks in advance!

Answer 3

@imqwerty

You helped me last time! Hope you don't mind the ping. ;u;

Answer 4

if u have something like this-
a+b=c
and if u take log on both sides u get this-
\[\log(a+b)=logc\]and not this-\[\log(a)+\log(b)=\log(c)\]

Answer 5

i'll advice u to break \[3^{2x+1}\]like this-\[(3^x)^2(3)\]and then u substitute \[3^x=a\]in whole equation
you will be left with a quadratic equation in a
u get 2 values of a
then u know a=3^x
then u find x :)

feel free the ask anytime (:

Answer 6

@imqwerty

Ah yes, I remember reading that I could substitute this stuff for a quadratic equation, I never remember to though, haha. Thanks for the help! This clears things up. ;u;

Answer 7

(: no prblm