How do you solve these logarithmic/exponent problems?

For math class, I have to solve pre-calculus problems during the summer, and there's three I don't understand.

2^x +4^x = 2^(x+1) - 4^(x+1)

ln(x) = 2-3(ln(x))

2log 9x = log (x+8) + 2

What do I do?
Thanks!

1) Rearrange terms to get powers of 2 on one side and powers of 4 on the other side, use laws of exponents, and use the fact that 2^2 = 4:

2^x + 4^x = 2^(x+1) - 4^(x+1)
4^x + 4^(x+1) = 2^(x+1) - 2^x
4^x + 4^x*4^1 = 2^x*2^1 - 2^x
4^x * (1 + 4) = 2^x * (2 - 1)
4^x * 5 = 2^x
(2^2)^x * 5 = 2^x
2^(2x) * 5 = 2^x
log [2^(2x) * 5] = log (2^x)
2x + log 5 = x
x = - log 5, or equivalently log (1/5).

2) Isolate the logarithm, solve for the logarithm, then solve for x:

ln(x) = 2-3(ln(x))
ln(x) + 3ln(x) = 2
4ln(x) = 2
ln(x) = 2/4 = 1/2
x = e^(1/2).

3) Get the logs on one side, use laws of logarithms to condense into one logarithm, then use the definition of a logarithm:

2log 9x = log (x+8) + 2
2log 9x - log (x+8) = 2
log (9x)^2 - log (x+8) = 2
log [(9x)^2 / (x + 8)] = 2
(9x)^2 / (x + 8) = 3^2 = 9
81x^2 / (x + 8) = 9
9x^2 / (x + 8) = 1
9x^2 = x + 8
9x^2 - x - 8 = 0
(x - 1)(9x + 8) = 0
x = 1 or x = -8/9

We can only take logs of positive numbers, so 9x and x + 8 must both be positive.
So we must discard x = -8/9 since this value of x would make 9x negative.

The solution set is x = 1 only.

Lord bless you today!

Question 1)
... 2^(x) + 4^(x) = 2^(x+1) - 4^(x+1)
or 2^(x) + 2^(2x) = 2 * 2^(x) - 4 * 2^(2x)
or 5 * 2^(2x) = 2^(x)
or ln(5) + 2x ln(2) = x ln(2)
or (2x - 1) ln(2) = - ln(5)
or x = (1/2) (1 - ln(5)/ln(2) )
or x ~ -0.661

Question 2)
... ln(x) = 2 - 3ln(x)
or 4ln(x) = 2
or ln(x) = 1/2
or x = e^(1/2)
or x ~ 1.649

Question 3)
... 2log3(9x) = log3(x+8) + 2
or log3( (9x)^2 ) = log3(x+8) + log3( 3^2 )
or log3( (9x)^2 ) = log3( 9(x+8) )
or 81x^2 = 9x + 72
or 9x^2 - x - 8 = 0
or (x - 1)(9x + 8) = 0
or x = -8/9 ← ignore, there's no such thing as log3(-8)
or x = 1 ← final answer

Please post only 1 question per post.

2^x + 4^x = 2^(x + 1) - 4^(x + 1)

2^(x) + 4^(x) = 2*2^(x) - 4*4^(x)

1 + 2^(x) = 2 - 4*2^(x)

5*2^(x) = 1

x = log(base 2) 1/5