How do you solve this logarithmic equation?

Question

logv3 x + logv3(x^2-8) = logv3 8x

the "v" means it is a little 3 under the log

Answer 1

X=8

X=-1

Answer 2

log3(x)+log3(x^2-8)-log3(8x)=0

x+x^2-8-8x=0

x^2-7x-8=0

(x-8)(x+1)=0

x=8,-1

cant be -1 because of logs

x=8

make it a good day

Answer 3

To do these log problems you need to memorize these properties:

log A + log B = log AB

log A - log B = log A/B

n log A = log A^n

and

b^log vb N =N

In your case:

logv3 x + logv3(x^2-8) = logv3 8x

logv3 [x*(x^2-8)] = logv3 8x

[x*(x^2-8)] =8x

x^3 -8x = 8x

x^3 - 16x = 0

x(x^2-16) = 0

x=0 x = 4, x = -4

Now since this is logs the domain must be positive. There are no logs of negatives or zero.

So x ≠ 0, -4

x =+ 4 only

Answer 4

Use the rule log(a) + log(b) = log(ab) with any base to combine the two on the left.

log x + log(x^2 - 8) = log x(x^2 - 8) = log 8x

x(x^2 - 8) = 8x

Can you finish it from there?

Edit. I intended to leave it to you but in view of the wrong answer above I will finish it.

x(x^2 - 8) = 8x

x^3 - 8x = 8x

x^3 - 16x = 0

x(x^2 - 16) = 0

x(x - 4)(x + 4) = 0

x = 0 or 4 or -4

However check with the original equation.

log(-4) and log(0) are undefined so x = -4 and x = 0 are not solutions.

This leaves x = 4

The most important thing to learn is that log(a) + log(b) IS NOT log(a + b)

It pains me to write that even with the words "is not" in there.

Answer 5

logv3  [ (x )(x^2-8)] = logv3 8x

[ (x )(x^2-8)] = 8x

(x^2-8) = 8

x^2=16

x=4