Question

log2(x+5)=4+log2(x-1)

Solve for x.

Answer 1

log2(x - 4) + log2(x) = 5 : x > 4
log2[x(x-4)] = 5
x(x-4) = 2^5 = 32
x² - 4x - 32 = 0
(x+4)(x-8) = 0
x = -4, 8 : but since x > 4, discard the extraneous x = -4 result

x = 8

Answer 2

Subtract log2(x-1) from both sides.

Answer 3

that is not an option the options are
a. {-2/5}
b. { 2/5}
c. {7/5}
d. {-7/5}

Answer 4

@Mertsj I got log2(x+5)-log2(x-1)=4

Answer 5

Now use the law that says that
\[\log_{b}m-\log_{b} n=\log_{b}\frac{m}{n} \]

Answer 6

What did you get?

Answer 7

@Mertsj I dont know anything about a law? But what about the 4?

Answer 8

\[\log_{2} \frac{x+5}{x-1}=4\]

Answer 9

Now put it into exponential form. What is the base?

Answer 10

you lost me with expo form but the base is 2 right?

Answer 11

And the log is the exponent so what is the exponent?

Answer 12

So you don't know the laws of logs and you don't know the exponential form of a log equation so why are you doing this problem? What are you supposed to be studying?

Answer 13

I know the exponential form of a log but I dont know how to do it for this equation. So do I move the 4 over or not?

Answer 14

The log is the exponent. So what is the exponent?

Answer 15

I don't know is it 1?