Question

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

Answer 1

`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`

Answer 2

`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`