# logarithms and domain, logarithm equations

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• Apr 13th 2009, 03:17 PM
jlefholtz
logarithms and domain, logarithm equations
I am not sure how to work these two kinds of problems out:

1) Solve this equation:

log2 (x-3)(x+1)=5

I keep getting x= 4 or x=-2 but the solution says that the answer is -5, 7. Could someone inform me where I have made a mistake?

log2 (x-3)(x+1)=5
log2 x^2-2x-3=5
log2 x^2-2x-8=0
log2 (x-1)^2-9
log2 (x-1+3)(x-1-3)
log2 (x+2)(x-4)
x=-2, x=4

2)Find the domain of the following:

1/log2 (x^2-6x+8)

I'm not quite sure how to even go about this one.

Thank you.
• Apr 13th 2009, 03:22 PM
Jhevon
Quote:

Originally Posted by jlefholtz
I am not sure how to work these two kinds of problems out:

1) Solve this equation:

log2 (x-3)(x+1)=5

I keep getting x= 4 or x=-2 but the solution says that the answer is -5, 7. Could someone inform me where I have made a mistake?

log2 (x-3)(x+1)=5
log2 x^2-2x-3=5
log2 x^2-2x-8=0
log2 (x-1)^2-9
log2 (x-1+3)(x-1-3)
log2 (x+2)(x-4)
x=-2, x=4

and you found no problem with bring the 5 over the equal sign and then putting it inside the log?

\$\displaystyle \log_2(x - 3)(x + 1) = 5 \implies (x - 3)(x + 1) = 2^5\$

Quote:

2)Find the domain of the following:

1/log2 (x^2-6x+8)

I'm not quite sure how to even go about this one.

Thank you.
you require two things here. the denominator of the fraction cannot be zero, and what is being logged must be greater than zero

can you finish up?
• Apr 13th 2009, 03:24 PM
jlefholtz
oh geez. i'm new to logarithms, i completely forgot about that. thanks for your help with both of the questions, i can go on from here.
• Apr 14th 2009, 05:12 AM
stapel
Quote:

Originally Posted by jlefholtz
oh geez. i'm new to logarithms....

Logs are functions, just like the other functions you've worked with. Just as f(x) + 5 is not at all the same thing as f(x + 5), so also log(x) + 5 is not the same as log(x + 5). :D