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9^x-2^(x+1/2)=2^(x+7/2)-3^(2x-1), then solve for x.

Posted in Math, asked by aimro, 6 years ago. 3840 hits.

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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

log81/448=x

x=log81/448

Venkat Nagendra Thati
Venkat Nagendra Thati - 6 years ago
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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)

 

vinu omanan
vinu omanan - 6 years ago
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