Given $log_a 2 =4, log_a 3 =5$ , and $log_a 11 = 8$ , what is $log_a (33)/(2a^3)$ ?

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Answer 1 · 2018-08-12T05:50:32

$X=(log_a 33)/(2a^3)=39/8$

We know that ,

$log_a x=y<=>$

Here ,

$log_a 2=4=>2=a^4 and log_a 3=5=>3=a^5$

$i.e. color(blue)(a^4=2 and$

$a^5=3=>a(color(blue)(a^4))=3=>a(color(blue)(2))=3=>color(green)(a=3/2$

Now , $a^4=2=>a^3color(green)((a))=2=>a^3(color(green)(3/2))=2$

$:.a^3=4/3=>color(red)(2a^3=8/3$

Let ,

$X=(log_a 33)/(2a^3)=(log_a(3xx11))/(8/3)to[:.2a^3=8/3]$

$:. X=(log_a3+log_a11)/(8/3)to[becauselog(m*n)$ = $logm+logn]$

$:.X=(5+8)/(8/3)=(13xx3)/8=39/8$

Given log_a 2 =4, log_a 3 =5log_a 2 =4, log_a 3 =5, and log_a 11 = 8log_a 11 = 8, what is log_a (33)/(2a^3)log_a (33)/(2a^3)?