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log4 6 + log4 (x + 1) = 2

Guest Apr 4, 2017

Best Answer 

 #2
avatar+79894 
+2

log4 6 + log4 (x + 1) = 2    

 

We can write

 

log4 [6 * (x + 1) ]  = 2

 

In exponential form, we have

 

4^2   =  6 (x + 1)    simplify

 

16 = 6x + 6      subtract 6 from both sides

 

10  = 6x      divide both sides by 6

 

10/ 6  = x  =   5/3

 

 

cool cool cool

CPhill  Apr 4, 2017
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2+0 Answers

 #1
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+1

Solve for x:
(log(6))/(log(4)) + (log(x + 1))/(log(4)) = 2

Rewrite the left hand side by combining fractions. (log(6))/(log(4)) + (log(x + 1))/(log(4)) = (log(6) + log(x + 1))/(log(4)):
(log(6) + log(x + 1))/(log(4)) = 2

Multiply both sides by log(4):
log(6) + log(x + 1) = 2 log(4)

Subtract log(6) from both sides:
log(x + 1) = 2 log(4) - log(6)

2 log(4) - log(6) = log(4^2) + log(1/6) = log(16) + log(1/6) = log(16/6) = log(8/3):
log(x + 1) = log(8/3)

Cancel logarithms by taking exp of both sides:
x + 1 = 8/3

Subtract 1 from both sides:
Answer: | x = 5/3

Guest Apr 4, 2017
 #2
avatar+79894 
+2
Best Answer

log4 6 + log4 (x + 1) = 2    

 

We can write

 

log4 [6 * (x + 1) ]  = 2

 

In exponential form, we have

 

4^2   =  6 (x + 1)    simplify

 

16 = 6x + 6      subtract 6 from both sides

 

10  = 6x      divide both sides by 6

 

10/ 6  = x  =   5/3

 

 

cool cool cool

CPhill  Apr 4, 2017

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