9^x-2^(x+1/2)=2^(x+7/2)-3^(2x-1)
we can rewrite the given equation with some changes
we can write 9^x in the form of 3^2x and also we will write a^m-a^n=a^m/a^n
so,3^2x/2^(x+1/2)=2^(x+7/2)/3^(2x-1)
= we will do cross multiplication in both the equations
= 3^2x*3^(2x-1)=2^(x+7/2)*2^(x+1/2)
we know that ,a^m*a^n=a^(m+n)
=> 3^(2x+2x-1)=2^(x+7/2+x+1/2)
= 3^(4x-1)=2^(2x+4)
=3^(4x-1)=4^(x+2)
=3^4x/3^1=4^x*4^2
=3^4x/4^x=16*3
=(81/4)^x=48
we can apply log here
log_{81/4}^{48}=x
x=log_{81/4}^{48}
9^x-2^(x+1/2)=2^(x+7/2)-3^(2x-1)
we can write 9^x=(3^2)^x=3^2x [since (a^m)^n=a^(m*n)]
3^(2x) - 2[x+(1/2)] = 2^[x+(7/2)] - 3^(2x-1)
on rearranging the terms,
3^(2x)+ 3^(2x-1) = 2^[x+(7/2)] + 2[x+(1/2)]
since a^(m+n) = a^m * a^n, and a^(m-n) = a^m / a^n
3^(2x) + [3^(2x)*3^(-1)] = [2^x *2^(7/2)] + [2^x * 2^(1/2)]
from two terms on LHS, we can take 3^2x as common and from two terms on RHS, we can take 2^x as common,
3^(2x)[1+1/3] = 2^x[2^(7/2) + 2^(1/2)]
3^2x[4/3] = 2^x[128^(1/2)+2^(1/2)]
3^2x[4/3] = 2^x[8*2^(1/2)+2^(1/2)]
3^2x*4 = 2^x * 9*2^(1/2)*3
(3^2x)/(2^x) = 27*2^(1/2)/4
(3^2x)/(2^x) = 27/2*2^(1/2)
(3^2x)/(2^x) = 27/2^(3/2)
(3^2x)/3^2=(2^x)/2^(3/2)
(3^(2x-2)) = (2^(x-(3/2)))
on taking log,
(2x-2)log3 = [(2x-3)/2]log2
4(x-1) log3 = (2x-3) log 2
4(x-1)/(2x-3) = log2/log3
Let log2/log3=y
y = 4(x-1)/(2x-3)
(2x-3)y = 4x-4
2xy-3y=4x-4
4x-2xy = 4-3y
2x(2-y) = 4-3y
2x=(4-3y)/(2-y)
x = (4-3y)/(4-2y)
x=4-3log2/log3 / 4-2log2/log3
= (4log3-3log2)/ (4log3-2log2)