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1.) Find all \(x\) so that \($8^{x^2} \cdot 4^x = 2$\). Enter all the solutions, separated by commas.

 

2.) 

Problem:

Find all real solutions to


\(2 \log_2 (x + 5) = \log_2 (x - 9) + \log_2 (x + 53) + 1 \)
Enter all the solutions, separated by commas.

 Mar 16, 2019
 #1
avatar+102417 
0

8^(x^2) * 4^x  =  2

 

[ 2^3 }^(x^2) * (2^2)^x  =  2^1

 

2^(3x^2) * 2^(2x)  = 2^1

 

2^( 3x^2 + 2x)  = 2^1         the bases are the same......solve for the exponents

 

3x^2 + 2x   = 1

 

3x^2 + 2x - 1  = 0

 

(3x - 1)(x +1) = 0

 

Setting both factors to 0   and solving for x produces  

 

x = 1 / 3    or   x   = -1

 

cool cool cool

 Mar 16, 2019
 #2
avatar+102417 
0

Note : 1  = log2(2)

 

2log2 ( x + 5) = log2(x - 9) + log2(x + 53) + log2(2)

 

log2(x + 5)^2  = log2(x - 9) + log2(x + 53) + log2(2)

 

log2(x + 5)^2  =  log2 [ (x - 9)(x + 53) * 2 ]        ....so...

 

(x + 5)^2  =  2(x - 9)(x + 53)

 

x^2 + 10x + 25  =  2 ( x^2 + 44x - 477)

 

x^2 + 10x + 25  = 2x^2 + 88x - 954

 

x^2 + 78x - 979 = 0

 

(x - 11) ( x + 89) = 0

 

Setting each factor to 0 and solving for x produces

 

x = 11  (accept)            or

 

x = -89  (reject as it doesn't give a real valued log  )

 

 

 

cool cool cool

 Mar 16, 2019

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