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

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: solve for x log5(4x)=log5 28 2log x+ log3= log48 2^2x+3 = 2^x-5 4^x= 28      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first two to get you started

# 1

log%285%2C%284x%29%29=log%285%2C%2828%29%29 Start with the given equation


4x=28 Since the logs have the same base, this means that the arguments (the stuff inside the logs) are equal.


x=%2828%29%2F%284%29 Divide both sides by 4 to isolate x



x=7 Divide

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Answer:
So our answer is x=7




# 2

2log%2810%2C%28x%29%29%2B+log%2810%2C%283%29%29=log%2810%2C%2848%29%29+ Start with the given equation


log%2810%2C%28x%5E2%29%29%2B+log%2810%2C%283%29%29=log%2810%2C%2848%29%29+ Rewrite 2log%2810%2C%28x%29%29 as log%2810%2C%28x%5E2%29%29


log%2810%2C%28x%5E2%2A3%29%29=log%2810%2C%2848%29%29+ Combine the logs using the identity log%28b%2C%28A%29%29%2Blog%28b%2C%28B%29%29=log%28b%2C%28A%2AB%29%29


log%2810%2C%283x%5E2%29%29=log%2810%2C%2848%29%29+ Rearrange the terms


3x%5E2=48 Since the logs have the same base, this means that the arguments (the stuff inside the logs) are equal


x%5E2=16 Divide both sides by 3


x=4 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 x=4