What is the answer to ((log_{2}(9-2^x)))/((3-x))=1 ?

The answer to ((log_{2}(9-2^x)))/((3-x))=1 is x=0

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((log_{2}(9-2^x)))/((3-x))=1

\frac{( lo g _{2} ( 9 - 2 ^{x} ) )}{( 3 - x )} = 1

x = 0

Solution steps

\frac{( lo g _{2} ( 9 - 2 ^{x} ) )}{( 3 - x )} = 1

Multiply both sides by

3 - x

lo g_{2} (9 - 2^{x}) = 3 - x

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

2^{l o g_{2} (9 - 2^{x})} = 2^{3 - x}

Simplify

2^{l o g_{2} (9 - 2^{x})} : 9 - 2^{x}

9 - 2^{x} = 2^{3 - x}

Solve

9 - 2^{x} = 2^{3 - x} : x = 0, x = 3

x = 0, x = 3

Verify Solutions:

x = 0

True

, x = 3

False

The solution is

x = 0

lo g_{2} (x) + 20 = 2

lo g_{4} (9 x - 5) = 3

lo g_{2} (x + 3) - lo g_{2} (x - 5) = 3

2 \cdot lo g_{3} (x) - lo g_{3} (x + 6) + 1 = 2

lo g_{2} (x) = 0

What is the answer to ((log_{2}(9-2^x)))/((3-x))=1 ?
The answer to ((log_{2}(9-2^x)))/((3-x))=1 is x=0
Solve for x: ((log_{2}(9-2^x)))/((3-x))=1
The solution is x=0