Respuesta :

The value of 'x' is 4

How to determine the value

Given that;

3^2x-2 -28(3^x-2) +3 = 0

Open the bracket, we have

3^2x-2 - 84^x-2 +3^1 = 0

Note that 81 which is within 84 is written in index form as 3^4, represent 84 with it and add 3 remaining

3^2x-2 - 3^4(x-2) + 3^1 + 3^1 = 0

Since we have all the numbers in index form, divided through and equate the numerators

2x-2 -4(x-2) + 1 + 1 = 0

Expand the bracket

2x - 2 - 4x + 8 +2 = 0

Collect like terms

2x- 4x + 8 = 0

-2x = -8

x = -8/-2

x= 4

Thus, the value of 'x' is 4

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Answer:

x = 3, x = 0

Given expression:

[tex]3^{2x-2}-28\left(3^{x-2}\right)+3=0[/tex]

rewrite expression

[tex]3^{2x} \times 3^{-2}-28\left(3^{x} \times 3^{-2}\right)+3=0[/tex]

treat [tex]\sf \s3^{x}\:as\:u[/tex], then expression

[tex]u^2 \times 3^{-2}-28\left(u \times 3^{-2}\right)+3=0[/tex]

multiply both sides by [tex]\sf 3^{2}[/tex]

[tex]u^2-28\left u+27=0[/tex]

factor expression

[tex]u^2-27\left u - u + 27=0[/tex]

factor out common terms

[tex]u(u-27) -1( u - 27)=0[/tex]

collect into groups

[tex](u-1)( u - 27)=0[/tex]

set to zero

[tex]u =1, \ u=27[/tex]

Substitute [tex]\sf 3^x = u[/tex], to find x

[tex]3^x = 1, \ 3^x = 27[/tex]

[tex]3^x = 3^0, \ 3^x =3^3[/tex]

[tex]x= 0, \ x = 3[/tex]

Find more on exponential equation's: brainly.com/question/27893440

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