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Suppose that a, b and are positive real numbers such that \(a^{\log_3 7} = 27\), \(b^{\log_7 11} = 49\), and \(c^{\log_{11}25} = \sqrt{11}\). Find the value of \(a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}\)
 

 Jan 2, 2019
 #1
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By the change-of-base theorem.....

log3 7  =  log 7 / log 3

So

a^(log 7 /log 3) = 27

So

a^(og3 7)^2 =

a^ [ (log 7/log3) * (log7 /log3 ) ] = 27^(log7/log3)

 

Similarly

b^[ log 11/ log 7]^2    =

b^[ (log 11/log 7) *(log11/log 7) ] = 49^(log11/log 7)

 

And

c^[(log25/log11) * (log25/log11)]  =   sqrt(11)^(log25/log11)

 

So

27^(log7/log3) + 49^(log11/log7) + sqrt(11)^(log 25/log11)   =

 

27^(log3 7) + 49^(log7 11)  +  sqrt (11)^(log11 25)  =

 

(3^3)^(log 3 7)  +  (7^2)^(log 7 11 ) +  (11^[1/2] )^(log11 25)  =

 

(3 ^log3 7)^3  + (7 ^ log 7 11)^2  +   (  11^log11 25 )^(1/2)

 

By a log property.....a^(log a b )  = b

 

So we have

 

(7)^3    + 11^2   +  25^(1/2)  =

 

343   +  121 +  5  =

 

469

 

 

cool cool cool

 Jan 2, 2019
edited by CPhill  Jan 2, 2019

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