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

Question

1228 views around the world

You can reuse this answer
Creative Commons License

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#