Question

{-243×3^x+5 +(3^x)^2}/3^2 ×3^x -3^12 =3^6x-62​

Answer 1

x = (7/2) + log3[1 + (3^7+243)^(-1/2)] or x = (7/2) + log3[1 - (3^7+243)^(-1/2)]

Given:

{-243×3^x+5 +(3^x)^2}/3^2 ×3^x -3^12 =3^6x-62​

To Find:

Simplify the equation

Solution:

The given equation is:

{-243×3^x+5 +(3^x)^2}/(3^2 ×3^x) - 3^12 = 3^(6x-62)

Simplifying the left-hand side of the equation:

{-243×3^x+5 +(3^x)^2}/(3^2 ×3^x) - 3^12 = (3^x/3^x)^2 - 3^12

{-243×3^x+5 +(3^x)^2}/(3^2 ×3^x) - 3^12 = (3^(2x) - 3^12)

{-243×3^x+5 +(3^x)^2}/(3^2 ×3^x) - (3^(2x) - 3^12) = 0

Now, we can substitute a variable, let's say y = 3^x.

Therefore, the equation can be rewritten as:

{-243y+5 + y^2}/(3^2y) - (y^2 - 3^12) = 0

Multiplying both sides of the equation by 3^2y:

{-243y+5 + y^2} - 3^14y = 0

y^2 - 243y + 5 - 3^14y = 0

y^2 - (243+3^14)y + 5 = 0

Applying the quadratic formula:

y = [{(243+3^14) ± sqrt((243+3^14)^2 - 4(1)(5))}/2(1)]

y = [{(243+3^14) ± sqrt(59049+2(243)(3^14)+3^28-20)}/2]

y = [{(243+3^14) ± sqrt(3^28+2(243)(3^14)+59029)}/2]

y = [{(243+3^14) ± sqrt((3^14+243)^2 - 59029)}/2]

y = [{(243+3^14) ± sqrt((3^14+243- 243)(3^14+243+243))/2]

y = [{(243+3^14) ± sqrt(3^14(3^14+486))}/2]

y = [3^7/2(3^7+243) ± 3^7/2]

y = [3^7(1 ± (3^7+243)^(-1/2))]/2

Now, substituting back y = 3^x:

3^x = [3^7(1 ± (3^7+243)^(-1/2))]/2

Taking the logarithm with base 3 on both sides:

x = log3[3^7(1 ± (3^7+243)^(-1/2))]/2

x = (7/2) + log3[1 ± (3^7+243)^(-1/2)]

Therefore, the solutions are:

x = (7/2) + log3[1 + (3^7+243)^(-1/2)] or x = (7/2) + log3[1 - (3^7+243)^(-1/2)]

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