What is 1/(log_{2)(10)-log_{2}(9)} ?

The solution to 1/(log_{2)(10)-log_{2}(9)} is 1/(1+log_{2)(5/9)}

Popular >

simplify 1/(log_{2)(10)-log_{2}(9)}

simplify

\frac{1}{lo g _{2} ( 10 ) - lo g _{2} ( 9 )}

\frac{1}{1 + lo g _{2} ( \frac{5}{9} )}

Decimal Notation

6.57881 \dots

Solution steps

\frac{1}{lo g _{2} ( 10 ) - lo g _{2} ( 9 )}

lo g_{2} (10) = 1 + lo g_{2} (5)

lo g_{2} (9) = 2 lo g_{2} (3)

= \frac{1}{1 + lo g _{2} ( 5 ) - 2 lo g _{2} ( 3 )}

Apply log rule:

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

- 2 lo g_{2} (3) = - lo g_{2} (3^{2})

= \frac{1}{1 + lo g _{2} ( 5 ) - lo g _{2} ( 3 ^{2} )}

lo g_{2} (5) - lo g_{2} (3^{2}) = lo g_{2} (\frac{5}{9})

= \frac{1}{1 + lo g _{2} ( \frac{5}{9} )}

lo g_{10} (5 x + 1) - lo g_{10} (x)

lo g_{8} (\frac{f}{64})

2 ln (x) + ln (y)

lo g_{b} (\frac{m}{n})

ln (x) - ln (y) + ln (z)

What is 1/(log_{2)(10)-log_{2}(9)} ?
The solution to 1/(log_{2)(10)-log_{2}(9)} is 1/(1+log_{2)(5/9)}