Question

Log3 (4x+1) log3(13x-5)=2

Answer 1

assuming base 3,

log((4x+1)(13x-5)) = 2
(4x+1)(13x-5) = 3^2 = 9
52x^2 - 7x - 5 = 9

Now just solve the quadratic

Answer 2

let Log3 (4x+1) = a and
log3(13x-5) = b , where x > 5/13
then ab = 2 --> b = 2/a

3^a = 4x+1
x = (3^a - 1)/4

3^b = 13x-5
x = (3^b + 5)/13

(3^a - 1)/4 = (3^(2/a) + 5)/13

arggghhh, let's see what Wolfram says, (I changed the a to x, Wolfram did not like the a)
http://www.wolframalpha.com/input/?i=solve+%283%5Ex+-+1%29%2F4+%3D+%283%5E%282%2Fx%29+%2B+5%29%2F13

a = 1.3052
then 3^1.3052 = 4x+1
4x = 3^1.3052 - 1

<b>x = .98766171</b>


looks like I could just as well just plugged the original equation into Wolfram:
http://www.wolframalpha.com/input/?i=solve+Log3+%284x%2B1%29+log3%2813x-5%29%3D2

Where did you get this question?
What level of math is this?

Answer 3

mmmhhh,
I thought logA + logB = log(AB)

I suspect a typo in the original question.