What is 1/(log_{4)(6)}+1/(log_{9)(6)} ?

The solution to 1/(log_{4)(6)}+1/(log_{9)(6)} is (log_{9}(6)+log_{4}(6))/(log_{4)(6)log_{9}(6)}

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simplify 1/(log_{4)(6)}+1/(log_{9)(6)}

simplify

\frac{1}{lo g _{4} ( 6 )} + \frac{1}{lo g _{9} ( 6 )}

\frac{lo g _{9} ( 6 ) + lo g _{4} ( 6 )}{lo g _{4} ( 6 ) lo g _{9} ( 6 )}

Solution steps

\frac{1}{lo g _{4} ( 6 )} + \frac{1}{lo g _{9} ( 6 )}

Least Common Multiplier of

lo g_{4} (6), lo g_{9} (6) : lo g_{4} (6) lo g_{9} (6)

Adjust Fractions based on the LCM

= \frac{lo g _{9} ( 6 )}{lo g _{4} ( 6 ) lo g _{9} ( 6 )} + \frac{lo g _{4} ( 6 )}{lo g _{4} ( 6 ) lo g _{9} ( 6 )}

Apply the fraction rule:

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

= \frac{lo g _{9} ( 6 ) + lo g _{4} ( 6 )}{lo g _{4} ( 6 ) lo g _{9} ( 6 )}

x^{6} - y^{6}

\frac{8}{\frac{3}{2}}

∣ x + 8 ∣ = 12

4 (1 - x) = 9 (1 + 2 x) + 9

3 n - 5 = 19

What is 1/(log_{4)(6)}+1/(log_{9)(6)} ?
The solution to 1/(log_{4)(6)}+1/(log_{9)(6)} is (log_{9}(6)+log_{4}(6))/(log_{4)(6)log_{9}(6)}