What is the answer to log_{a}(x^c)=clog_{a}(x) ?

The answer to log_{a}(x^c)=clog_{a}(x) is True for all x

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log_{a}(x^c)=clog_{a}(x)

lo g_{a} (x^{c}) = c lo g_{a} (x)

True for all x

Solution steps

lo g_{a} (x^{c}) = c lo g_{a} (x)

Apply log rules

c lo g_{a} (x) = c lo g_{a} (x)

Rewrite the equation with

lo g_{a} (x) = u

c u = c u

Both sides are equal

True for all u

True for all x

4^{2} \cdot lo g_{4} (x + 2) = 32

- 3.5 = 5 lo g_{10} (x) - 5

lo g_{- 2 x - 2} (3 x^{2} - 4 x - 16) = 2

- \frac{ln ( - 10 ( 0 ) ^{2} - x )}{4} = 0

x, f ln (x) = 1

What is the answer to log_{a}(x^c)=clog_{a}(x) ?
The answer to log_{a}(x^c)=clog_{a}(x) is True for all x