Exponential Equation #2^(x+1) + 2^(x+2) + 3(2^(x-2)) = 8(3^(x-2))# ?

A step by step solution of the following would be greatly appreciated
Thank you :D

2 Answers
Jun 4, 2016

Answer:

#x=5#

Explanation:

Consider #2^(x+1) -> 2^(x)xx2#

Consider #2^(x+2)->2^(x)xx2^2#

Consider #3(2^(x-2)) -> 2^(x)xx3/(2^2)#

Consider #8(3^(x-2)) ->3^(x)xx8/3^2#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#" "2^x(2+2^2+3/2^2) = 8/3^2xx3^x#

#" "2^(x)xx27/4=8/9xx3^x#

Divide both sides by #3^x#

#" "2^(x)/(3^x)xx27/4=8/9xx3^x/(3^x)#

But #(3^x)/(3^x)=1#

#" "2^(x)/(3^x)xx27/4=8/9#

Multiply both sides by #4/27#

#" "2^(x)/(3^x)xx27/4xx4/27=8/9xx4/27#

But #27/4xx4/27 =27/27xx4/4 = 1#

#" "2^(x)/(3^x)=8/9xx4/27#

#" "2^x/3^x = 32/243#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From this both should be true that

#" "2^x=32" and "3^x=243#

Using logs

#" "xln(2)=ln(32) => x= ln(32)/ln(2)#

#" "=> x=5#
'..........................................

#" "xln(3)=ln(243)=> x=ln(243)/(ln(3)#

#" "=>x= 5#

Jun 4, 2016

Answer:

#x=5#

Explanation:

#2times 2^x+4 times2^x+3/4 times 2^x=(8/9)3^x#

grouping

#(27/4)2^x=(8/9) 3^x#
#(3/2)^x=(27times9)/(8times 4)=(3/2)^5#

#x = 5#