LOGARITHMS help please...?

Question

please walk make through these...

1/8=4^(6x-5)

and

log2 X + log2 (X+6) = 4

Answer 1

1/8 = 2^(-3)
4 = 2^2

So 2^(-3) = (2^2)^(6x-5)

2^(-3) = 2^(12x-10) (power of a power: multiply the powers)

-3 = 12x - 10
7 = 12x
x = 7/12 *******


Other prob:

Left side: log 2 (x^2 + 6x) (To add logs: multiply the part after the log)

so log 2 (x^2 + 6x) = 4

Now change to an exponential eqn:

2^4 = x^2 + 6x

16 = x^2 + 6x

x^2 + 6x - 16 = 0

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

x=-8 or x= 2

But -8 is not possible, so x = 2 ***********

Answer 2

First remember the three properties of logs.
logA+logB=log(AB)
logA-logB=log(A/B)
logA^B=BlogA
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Number 1

1/8=4^(6x-5)
Take ln of both sides ln(1/8)=ln(4^(6x-5))
Use the third property ln(1/8)=(6x-5)(ln4)
Solve the logs with calculator -2.0794=(6x-5)(1.3863)
Now just solve the problem -2.0794=8.3178x-6.9315
4.8521=8.3178x
x=.5833
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Number 2

log2 (x) + log2 (x+6) =4
Use the first property: log2(x^2+6x)=4
Make both sides powers above two. This will eliminate the log2
2^(log2(x^2+6x))=2^4
x^2+6x=16
x^2+6x-16=0
Factor to (x-2)(x+8)
Solutions are 2 and -8. Plug in to check.

-8 does not work because you cannot take the log of a negative number!!!
2 works (1+3=4)

Solution is 2.

Answer 3

2^-3 = 2^(12x-10)

-3 = 12x - 10

12x = 7

x = 7/12

log2 {x*(x+6)} = 4

x^2 + 6x = 2^4 = 16

x^2 + 6x -16 = 0

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

x = 2
x = -8