What is the answer to log_{4}(2^x+48)=x-1 ?

The answer to log_{4}(2^x+48)=x-1 is x=4

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log_{4}(2^x+48)=x-1

lo g_{4} (2^{x} + 48) = x - 1

x = 4

Solution steps

lo g_{4} (2^{x} + 48) = x - 1

f (x) = g (x),

then

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

4^{l o g_{4} (2^{x} + 48)} = 4^{x - 1}

Simplify

4^{l o g_{4} (2^{x} + 48)} : 2^{x} + 48

2^{x} + 48 = 4^{x - 1}

Solve

2^{x} + 48 = 4^{x - 1} : x = 4

x = 4

Verify Solutions:

x = 4

True

The solution is

x = 4

lo g_{4} (x^{2} + 3 x - 1) = lo g_{4} (5 x - 1)

lo g_{x} (126) = 3

ln (2 x) = ln (x + 2)

lo g_{10} (10 x) = 3

lo g_{10} (10 x) = 2

What is the answer to log_{4}(2^x+48)=x-1 ?
The answer to log_{4}(2^x+48)=x-1 is x=4
Solve for x: log_{4}(2^x+48)=x-1
The solution is x=4