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

The solution to log_{2}(x+1)-1log_{2}(2x) is log_{2}((x+1)/x)-1

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simplify log_{2}(x+1)-1log_{2}(2x)

simplify

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

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

Solution steps

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

Apply log rule:

b \cdot lo g_{a} (x) = lo g_{a} (x^{b})

- 1 \cdot lo g_{2} (2 x) = - lo g_{2} ((2 x)^{1})

= lo g_{2} (x + 1) - lo g_{2} ((2 x)^{1})

Apply log rule:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

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

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

Simplify

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

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

Apply log rule:

lo g_{a} (\frac{x}{y}) = lo g_{a} (x) - lo g_{a} (y)

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

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

Simplify

lo g_{2} (2 x) : 1 + lo g_{2} (x)

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

Apply the distributive law:

- (a + b) = - a - b

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

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

Apply log rule:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

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

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

n^{2} + 100 n^{2} + 1. 1^{n}

\frac{x ^{2}}{0.0032}

8 x^{2} y - 4 x y^{2}

3 lo g_{10} (x) - lo g_{10} (y) + 2 lo g_{10} (z)

\frac{( 4 x ^{2} \cdot y ^{2} ) ^{2}}{( 2 x ^{2} \cdot y ^{2} )}

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