What is the answer to k=log_{2}(k)+1 ?

The answer to k=log_{2}(k)+1 is k=2,k=1

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k=log_{2}(k)+1

k = lo g_{2} (k) + 1

k = 2, k = 1

Solution steps

k = lo g_{2} (k) + 1

Switch sides

lo g_{2} (k) + 1 = k

Move

1

to the right side

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

f (x) = g (x),

then

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

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

Simplify

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

k = 2^{k - 1}

Solve

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

k = 2, k = 1

Verify Solutions:

k = 2

True

, k = 1

True

The solutions are

k = 2, k = 1

lo g_{\frac{1}{2}} (x) = 1024

ln (2 x + 5) = 0

lo g_{10} (a) + lo g_{a} (100) = 3

ln (lo g_{10} (ln (x))) = 0.2

lo g_{\frac{1}{2}} (4 - 2 x) + 2 = 0