Question

Can someone solve this and show work? thanks.

Given logb a = x , then bx = a where
b > 0 but b #1, and a > 0.

1. 3log2 x = 12 2. log5 125 = x
2. 3 + 4 logx 4 = 5

Answer 1

\[3\log_{2} X = 12 --- 3+4\log_{x} 4=5 --- \log_{5} 125=x\]

Answer 2

\[3log_2x = 12 \]Divide both sides by 3:\[log_2x = 4 \implies 2^4 = x = 16\]

Now you try the rest.

Answer 3

And I'll check your answers, or help if you get stuck.

Answer 4

So the next problem is:

\[log_5(125) = x\]

Right? How can you re-write that?

Answer 5

5^x = 125

Answer 6

That is exactly right. So we need to know what power of 5 is 125. That power will be the x

Answer 7

So 125 is which power of 5?

Answer 8

3

Answer 9

Correct. So if:

\[5^x = 5^3\]
What must x = ?

Answer 10

3?
not sure.

Answer 11

Yep. 5^x = 5^3 means x = 3