What is the answer to log_{2}(n)=n-1 ?

The answer to log_{2}(n)=n-1 is n=2,n=1

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log_{2}(n)=n-1

lo g_{2} (n) = n - 1

n = 2, n = 1

Solution steps

lo g_{2} (n) = n - 1

f (x) = g (x),

then

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

2^{l o g_{2} (n)} = 2^{n - 1}

Simplify

2^{l o g_{2} (n)} : n

n = 2^{n - 1}

Solve

n = 2^{n - 1} : n = 2, n = 1

n = 2, n = 1

Verify Solutions:

n = 2

True

, n = 1

True

The solutions are

n = 2, n = 1

lo g_{2} (x) + lo g_{2} (2 x + 3) = 1

ln (5) - ln (4 - 4 x) = ln (33)

\frac{ln ( x )}{x} x = 1

lo g_{2} (x - 10) = 5

63 = 95 - 93 lo g_{10} (t + 1)