Question

help please !

Answer 1

\[\log_{3} (x-2)+\log_{3} (x-8)=\log_{3} ((x-2)(x-8))\]

Answer 2

clear now?

Answer 3

\[\log_{3} ((x-2)(x-8))=3\]
\[(x-2)(x-8)=3^3\]
\[(x-2)(x-8)=27\]
\[x^2-10x+16=27\]
\[x^2-10x+16-27=0\]
\[x^2-10x-11=0\]

Answer 4

now solve for x

Answer 5

First, combine both logs. You can do so since they have the SAME BASE. So:
\[(x-2)(x-8)= x ^{2}-10x+16\]
So, you have:
\[\log_{3} (x ^{2}-10x+16)= 3\]
One of the logarithm rules states:
\[\log _{a}(x)=y\] can be turned into:
\[a ^{y}=x\]
y is 3
a is also 3
x is (x ^{2}-10x+16)
Therefore:
\[3^{3}= x ^{2}-10x+16\]
This can be simplified to:
\[27=x ^{2}-10x+16\]
Subtract both sides by 27, and we get:
\[x ^{2}-10x-11=0\]
Solving this quadratic, we get:
x=11
x=-1
You can't have the log of a NEGATIVE number. Therefore, -1 is NOT your answer. The only answer choice left is 11. So, x=11. :D

Answer 6

>>You can't have the log of a NEGATIVE number.

@ags2658 Why can't we have the log of a negative number? What about the log of zero?

Answer 7

Well, think about it this way.
For example:
\[\log _{2}(-8)\]
Putting it into exponential form, we get:
\[2^{x}=-8\]
There is not a single real number for ''x'' that can get us -8.
This is why you can't find the log of a negative number.
The log of 0 is also UNDEFINED. You can't get 0 by raising anything to another number. :D

Answer 8

im confused lol z

Answer 9

still need help?

Answer 10

yes.

Answer 11

@Directrix please explain to me

Answer 12

>>please explain to me

The part about not taking the log of a negative number or the question you posted? @marcelie

Answer 13

the question i posted