Question

log_14(6-x)+log_14(1-x)=1
Solve exactly for X.

Answer 1

so, log_14(x^2-7x+6)=1
because of loga+logb=log(ab)
So what do i do with the One?

Answer 2

Write 1 as a log base 14.

Answer 3

How? XD

Answer 4

Logs arent my fav part of math lol, and we sorta went through this quickly ^_^"

Answer 5

Remember that a log is an exponent.
For example, the log base 10 of 100 is 2.
That means when you raise the base, 10, to the log (an exponent) of 2, you get 100.

Answer 6

We need to write 1 as a log base 14.

\(\log_{14} x = 1\)

14 is the base.
1 is the log, so it's the exponent.
x is the number which you get when you raise the base to the log (the exponent).

We need to find x.
Look at the this rule:

\(\log_b x = y~~~\iff~~~b^y = x\)

Answer 7

When going from a log expression to an exponential expression, the base remains the base.
The log is the exponent.
The number whose log you took, is equal to the base raised to the exponent (the log).

Answer 8

so then X would equal 14...
ahhh

Answer 9

Exactly.

Answer 10

So now you write 1 as the log base 14 of 14.

Answer 11

So... log_14(x^2-7x+6)=log_14(14)
so x^2-7x+6=14
so log x^2-7x-8=0
quadratic formula...

Answer 12

or factor..

Answer 13

x+1)(x-8

Answer 14

\(\log_{14} (6-x) + \log_{14} (1-x) = 1\)

\(\log_{14} [(6-x)(1-x)] = \log_{14} 14\)

\((6 - x)(1 - x) = 14\)

\(6 - 6x -x + x^2 = 14\)

\(x^2 -7x - 8 = 0\)

I got the same you did.

Answer 15

Yes, you can factor.

Answer 16

x=-1 is the answer?

Answer 17

because x=8 gives us negative answers in the log.

Answer 18

\((x - 8)(x + 1) = 0\)

\(x = 8 \) or \(x = -1\)

Now you need to check both answers in the original equation to make sure there are no problems with the logs. (Logs, square roots, denominators, etc. can create domain problems.)

Answer 19

You are exactly correct.
8 is eliminated, and only x = -1 is the solution.

Answer 20

Good job!

Answer 21

thank you

Answer 22

You're welcome.