- #1

Mootjeuh

- 5

- 0

## Homework Statement

I'm trying to solve trying logarithmic and exponential equations but for the life of me I can't seem to figure out what I'm doing wrong.

## Homework Equations

Log equation: log

_{3}(x+2)+1 = log

_{3}x²+4x

Exp equation: 2

^{x}+ 2

^{x+1}= 3

## The Attempt at a Solution

For the first one, here's what I did:

- 1 = log
_{3}(x²+4x) - log_{3}(x+2) - log
_{3}((x²+4x)/(x+2)) = 1 - log
_{3}((x**²**+4x)/(**x**+2)) = 1 - log
_{3}((x+**4**x)/**2**) = 1 --Can I do this? Dividing the denominator and numerator by the same value within a log - log
_{3}(x+2x) = 1 - log
_{3}(3x) = 1 - log
_{3}3 + log_{3}x = 1 - 1 + log
_{3}x = 1 - log
_{3}x = 0 - x = 1

And then for the second one:

- 2
^{x}+ 2^{x+1}= 3 - I figured I'd put them all on the same base so I tried to find 2 to what power equals 3
- So I did ln(3)/ln(2) = 1.584962501, which for simplicity's sake I'll replace by
*y*here - And when I put in 2
^{y}in my calculator it does give me 3 so it must be correct - x + x + 1 = y
- 2x + 1 = y
- x = (y-1)/2

I'm at my wit's end here, I don't know what else to try.

Any help is greatly appreciated.