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Find x if log_2 x^2 + log_1/2 x = 5.

 Feb 22, 2018
 #1
avatar+99586 
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log2 x^2 + log 1/2 x   =  5

 

2 log2 x + log 1/2 x  =  5            use the change-of-base rule to write

 

2 [ log x / log 2]  +  log x / log (1/2)  =  5        { log (1/2)  =  log 2^(-1) }

 

2[log x ] / log 2 ] + log x / log 2^(-1) =  5

 

2[;og x / log 2 ] +  log x / -log 2  = 5

 

2 [ log x / log 2 ]  - logx/ log 2  = 5

 

[2logx - log x ] / log 2  = 5

 

log x/ log 2  = 5

 

log x  =  5log 2

 

log x  =  log 2^5

 

log x  = log 32

 

x  = 32

 

 

cool cool cool

 Feb 22, 2018
 #2
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Solve for x:

(log(x^2))/log(2) - log(x)/log(2) = 5

 

Rewrite the left hand side by combining fractions. (log(x^2))/log(2) - log(x)/log(2) = (log(x^2) - log(x))/log(2):

(log(x^2) - log(x))/log(2) = 5

 

Multiply both sides by log(2):

log(x^2) - log(x) = 5 log(2)

 

log(x^2) - log(x) = log(1/x) + log(x^2) = log(x^2/x) = log(x):

log(x) = 5 log(2)

 

5 log(2) = log(2^5) = log(32):

log(x) = log(32)

 

Cancel logarithms by taking exp of both sides:

x = 32

 Feb 22, 2018

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