Question

log base 7 (x+4) + log base 7 (x-2) =1

Answer 1

exact same idea as before

Answer 2

i know it just didn't work out for me this time

Answer 3

1) combine into one log
\[\log_7((x+4)(x-2))=1\]

Answer 4

2)rewrite in equivalent exponential form:
\[(x+4)(x-2)=7\]

Answer 5

3) solve the resulting quadratic equation

Answer 6

discard any answers that would make the input of your logarithm negative

Answer 7

...

Answer 8

log base 7 (x+4) + log base 7 (x-2) =1
log base 7 (x+4)(x-2)=1
x^2+2x-8=7
x^2+2x-15=0
x^2+5x-3x-15=0
(x+5)(x-3)=0
x=-5,x=3

Answer 9

nvm i am having trouble concentrating! i get the questions sorry for the trouble of answering them.. maybe i should pay attention to what im writing

Answer 10

yeas and don't forget to discard -5 because it does not work!

Answer 11

Yeah I made a point about antilogarithms in your previous post. The always have to be greater than zero or the same way satellite put it you have to discard any solutions that make the input of your log zero. One way to do it is to set up the anti logs like this:
x + 4 > 0 , and x - 2 >0,

When you solve this x > 2 so when you get a solution like x = -5 it doesn't satisfy this condition and therefore that solution is not true.

Answer 12

want some medals?..
\[\log_{2}(2m+4)-\log_{2}(m-1)=3\]