Exponential Equation 2^(x+1) + 2^(x+2) + 3(2^(x-2)) = 8(3^(x-2))2x+1+2x+2+3(2x2)=8(3x2) ?

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

2 Answers
Jun 4, 2016

x=5x=5

Explanation:

Consider 2^(x+1) -> 2^(x)xx22x+12x×2

Consider 2^(x+2)->2^(x)xx2^22x+22x×22

Consider 3(2^(x-2)) -> 2^(x)xx3/(2^2)3(2x2)2x×322

Consider 8(3^(x-2)) ->3^(x)xx8/3^28(3x2)3x×832

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

" "2^x(2+2^2+3/2^2) = 8/3^2xx3^x 2x(2+22+322)=832×3x

" "2^(x)xx27/4=8/9xx3^x 2x×274=89×3x

Divide both sides by 3^x3x

" "2^(x)/(3^x)xx27/4=8/9xx3^x/(3^x) 2x3x×274=89×3x3x

But (3^x)/(3^x)=13x3x=1

" "2^(x)/(3^x)xx27/4=8/9 2x3x×274=89

Multiply both sides by 4/27427

" "2^(x)/(3^x)xx27/4xx4/27=8/9xx4/27 2x3x×274×427=89×427

But 27/4xx4/27 =27/27xx4/4 = 1274×427=2727×44=1

" "2^(x)/(3^x)=8/9xx4/27 2x3x=89×427

" "2^x/3^x = 32/243 2x3x=32243
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From this both should be true that

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

Using logs

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

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

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

" "=>x= 5 x=5

Jun 4, 2016

x=5x=5

Explanation:

2times 2^x+4 times2^x+3/4 times 2^x=(8/9)3^x2×2x+4×2x+34×2x=(89)3x

grouping

(27/4)2^x=(8/9) 3^x(274)2x=(89)3x
(3/2)^x=(27times9)/(8times 4)=(3/2)^5(32)x=27×98×4=(32)5

x = 5x=5