What is 4log_{2}(x+8)-log_{2}(x-9)-log_{2}(x-8) ?

The solution to 4log_{2}(x+8)-log_{2}(x-9)-log_{2}(x-8) is log_{2}(((x+8)^4)/((x-9)(x-8)))

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simplify 4log_{2}(x+8)-log_{2}(x-9)-log_{2}(x-8)

simplify

4 lo g_{2} (x + 8) - lo g_{2} (x - 9) - lo g_{2} (x - 8)

lo g_{2} (\frac{( x + 8 ) ^{4}}{( x - 9 ) ( x - 8 )})

Solution steps

4 lo g_{2} (x + 8) - lo g_{2} (x - 9) - lo g_{2} (x - 8)

Apply log rule:

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

4 lo g_{2} (x + 8) = lo g_{2} ((x + 8)^{4})

= lo g_{2} ((x + 8)^{4}) - lo g_{2} (x - 9) - lo g_{2} (x - 8)

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{2} ((x + 8)^{4}) - lo g_{2} (x - 9) = lo g_{2} (\frac{( x + 8 ) ^{4}}{x - 9})

= lo g_{2} (\frac{( x + 8 ) ^{4}}{x - 9}) - lo g_{2} (x - 8)

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{2} (\frac{( x + 8 ) ^{4}}{x - 9}) - lo g_{2} (x - 8) = lo g_{2} (\frac{\frac{( x + 8 ) ^{4}}{x - 9}}{x - 8})

= lo g_{2} (\frac{\frac{( x + 8 ) ^{4}}{x - 9}}{x - 8})

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= lo g_{2} (\frac{( x + 8 ) ^{4}}{( x - 9 ) ( x - 8 )})

ln (x - 1) - 2 ln (x) + ln (x + 1)

ln (2) \cdot \frac{1}{3}

5 ln (3) + ln (4)

\frac{ln ( 2.1 ) - ln ( 2 )}{0.1}

\frac{1}{4} ln (0 - 1) - \frac{1}{4} ln (0 + 3)

What is 4log_{2}(x+8)-log_{2}(x-9)-log_{2}(x-8) ?
The solution to 4log_{2}(x+8)-log_{2}(x-9)-log_{2}(x-8) is log_{2}(((x+8)^4)/((x-9)(x-8)))