Question

solve for x 9^x-3^x-8=0​

Answer 1

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First, you need to use the properties of exponents to rewrite 9^x as (3^x)^2, and 3^(x+1) as 3(3^x):

9^x = (3^2)^x = 3^(2x) = (3^x)^2 because (a^m)^n = a^(mn)

3^(x+1) = (3^x)(3^1) = (3^x)(3) = 3(3^x) because (a^m)(a^n) = a^(m+n)

Then your equation becomes (3^x)^2 - 3(3^x) - 54 = 0 and is seem to be quadratic in 3^x, suggesting the substitution u = 3^x:

u^2 - 3u - 54 = 0

This equation can be solved by factoring:

(u + 6)(u - 9) = 0

u + 6 = 0 or u - 9 = 0

u = -6 or u = 9

Now back-substituting,

3^x = -6 or 3^x = 9

The first equation does not have a solution in the real number system, because 3^x cannot be negative, but the second one does:

3^x = 9

3^x = 3^2

So, x = 2

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Answer 2

Answer:

First, you need to use the properties of exponents to rewrite 9^x as (3^x)^2, and 3^(x+1) as 3(3^x):

9^x = (3^2)^x = 3^(2x) = (3^x)^2 because (a^m)^n = a^(mn)

3^(x+1) = (3^x)(3^1) = (3^x)(3) = 3(3^x) because (a^m)(a^n) = a^(m+n)

Then your equation becomes (3^x)^2 - 3(3^x) - 54 = 0 and is seem to be quadratic in 3^x, suggesting the substitution u = 3^x:

u^2 - 3u - 54 = 0

This equation can be solved by factoring:

(u + 6)(u - 9) = 0

u + 6 = 0 or u - 9 = 0

u = -6 or u = 9

Now back-substituting,

3^x = -6 or 3^x = 9

The first equation does not have a solution in the real number system, because 3^x cannot be negative, but the second one does:

3^x = 9

3^x = 3^2

So, x = 2