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

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

anonymous · Answer

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

anonymous · Answer

$$3log_2x = 12 $$Divide both sides by 3:$$log_2x = 4 \implies 2^4 = x = 16$$

Now you try the rest.

anonymous · Answer

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

anonymous · Answer

So the next problem is:

$$log_5(125) = x$$

Right? How can you re-write that?

anonymous · Answer

5^x = 125

anonymous · Answer

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

anonymous · Answer

So 125 is which power of 5?

anonymous · Answer

3

anonymous · Answer

Correct. So if:

$$5^x = 5^3$$
What must x = ?

anonymous · Answer

3?
not sure.

anonymous · Answer

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