What is expand g(x)(x=2+log_{4}(x+1)) ?

The solution to expand g(x)(x=2+log_{4}(x+1)) is xg(x)=2+log_{4}(x+1)g(x)

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expand g(x)(x=2+log_{4}(x+1))

expand

g (x) (x = 2 + lo g_{4} (x + 1))

xg (x) = 2 + lo g_{4} (x + 1) g (x)

Solution steps

g (x) (x = 2 + lo g_{4} (x + 1))

= g (x) (2 = x + lo g_{4} (x + 1))

Apply the distributive law:

a (b + c) = ab + a c

a = g (x), b = x = 2, c = lo g_{4} (x + 1)

g (x) x = 2 + g (x) lo g_{4} (x + 1)

xg (x) = 2 + lo g_{4} (x + 1) g (x)

\frac{d}{d s} \frac{1}{( s ( s + 2 ) ^{2} )} \cdot (s + 2)

(2 x - y = 2,) (x + y + 3 = 0)

(23 - 32) (5 + 43)

\frac{s ^{2} + s + 1}{s ^{3} ( s + 1 )}

\frac{1}{( x ( x ^{2} + 1 ) )}

What is expand g(x)(x=2+log_{4}(x+1)) ?
The solution to expand g(x)(x=2+log_{4}(x+1)) is xg(x)=2+log_{4}(x+1)g(x)