What is expand log_{7}(2)(x+3)-log_{7}(x+1) ?

The solution to expand log_{7}(2)(x+3)-log_{7}(x+1) is log_{7}((2^{x+3})/(x+1))

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expand log_{7}(2)(x+3)-log_{7}(x+1)

expand

lo g_{7} (2) (x + 3) - lo g_{7} (x + 1)

lo g_{7} (\frac{2 ^{x + 3}}{x + 1})

Solution steps

lo g_{7} (2) (x + 3) - lo g_{7} (x + 1)

Expand

lo g_{7} (2) (x + 3) : lo g_{7} (2) x + lo g_{7} (2) \cdot 3

= lo g_{7} (2) x + lo g_{7} (2) \cdot 3 - lo g_{7} (x + 1)

Simplify

lo g_{7} (2) x + lo g_{7} (2) \cdot 3 - lo g_{7} (x + 1) : lo g_{7} (\frac{2 ^{x + 3}}{x + 1})

= lo g_{7} (\frac{2 ^{x + 3}}{x + 1})

(2 x^{\frac{2}{3}} + 3)^{2}

(t + 5)^{2} e^{a t}

(2 cos (x) - 1) (cos (x) + 1)

x (8 - 2 ln (x))

2 (1 + sin (θ))

What is expand log_{7}(2)(x+3)-log_{7}(x+1) ?
The solution to expand log_{7}(2)(x+3)-log_{7}(x+1) is log_{7}((2^{x+3})/(x+1))