Question

PLEASE HELP, I WILL MEDAL/FAN

Answer 1

2logx = log2 + log (3x-4)

Answer 2

Is this Calculus?

Answer 3

Solve for x:
2%2Alog%28%28x%29%29=log%28%282%29%29%2Blog%28%283x-4%29%29

Use this rule of exponents on the left side

A%2Alog%28%28B%29%29=log%28%28B%5EA%29%29

log%28%28x%5E2%29%29=log%28%282%29%29%2Blog%28%283x-4%29%29

Use this rule of logarithms on the right side:

log%28%28A%29%29+%2B+log%28%28B%29%29+=+log%28%28AB%29%29

log%28%28x%5E2%29%29=log%28%282%283x-4%29%29%29

Distribute on the right:

log%28%28x%5E2%29%29=log%28%286x-8%29%29%29

Use the rule:

The equation log%28%28A%29%29 = log%28%28B%29%29 is equivalent to
the equation A=B

x%5E2=6x-8

x%5E2-6x%2B8=0

%28x-2%29%28x-4%29=0

x-2=0 gives solution x=2

x-4=0 gives solution x=4

Answer 4

Sorry computer = slow -_-

Answer 5

Its pre calculus @Austin1617

Answer 6

nothing is showing up for what you wrote??

Answer 7

hold on

Answer 8

ok

Answer 9

So after you get the trinomial "\[x^2-6x+8\]" set it equal to zero then set each binomial equal to zero and: voila!

x-2=0 gives solution x=2

x-4=0 gives solution x=4

Answer 10

Remember:
log(A)+loh(B)=log(AB)