SOLUTION: solve for x log5(4x)=log5 28 2log x+ log3= log48 2^2x+3 = 2^x-5 4^x= 28

Question 143989: solve for x
log5(4x)=log5 28
2log x+ log3= log48
2^2x+3 = 2^x-5
4^x= 28
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'll do the first two to get you started

# 1

Start with the given equation

Since the logs have the same base, this means that the arguments (the stuff inside the logs) are equal.

Divide both sides by 4 to isolate x

Divide

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Answer:
So our answer is

# 2

Start with the given equation

Rewrite as

Combine the logs using the identity

Rearrange the terms

Since the logs have the same base, this means that the arguments (the stuff inside the logs) are equal

Divide both sides by 3

Take the square root of both sides. Note: discard the negative square root. Remember, you cannot take the log of a negative number.

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Answer:
So our answer is
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