What is the answer to (a)log_{3}(1+log_{3}(2^x-7))=1 ?

The answer to (a)log_{3}(1+log_{3}(2^x-7))=1 is x=(ln(3^{3^{1/a}-1}+7))/(ln(2))

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(a)log_{3}(1+log_{3}(2^x-7))=1

(a) lo g_{3} (1 + lo g_{3} (2^{x} - 7)) = 1

x = \frac{ln ( 3 ^{3^{\frac{1}{a}} - 1} + 7 )}{ln ( 2 )}

Solution steps

(a) lo g_{3} (1 + lo g_{3} (2^{x} - 7)) = 1

Rewrite the equation with

2^{x} = u

a lo g_{3} (1 + lo g_{3} (u - 7)) = 1

Solve

a lo g_{3} (1 + lo g_{3} (u - 7)) = 1 : u = 3^{3^{\frac{1}{a}} - 1} + 7 {a < 0 or a > 0}

u = 3^{3^{\frac{1}{a}} - 1} + 7 {a < 0 or a > 0}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 3^{3^{\frac{1}{a}} - 1} + 7 : x = \frac{ln ( 3 ^{3^{\frac{1}{a}} - 1} + 7 )}{ln ( 2 )}

x = \frac{ln ( 3 ^{3^{\frac{1}{a}} - 1} + 7 )}{ln ( 2 )}

2^{x - 3} = 128

(7^{x + 1})^{\frac{1}{5}} = 7

2^{x} + 1 + 2^{x} + 2 = 48

3^{x + 7} = 9^{x + 1}

x, 3^{x} + 2^{y} = 1 + 2^{z}

What is the answer to (a)log_{3}(1+log_{3}(2^x-7))=1 ?
The answer to (a)log_{3}(1+log_{3}(2^x-7))=1 is x=(ln(3^{3^{1/a}-1}+7))/(ln(2))