Show that `1/(log6 a) + 1/(log4 a) = 1/(log24 a)`

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We will use formula for change of base:

`log_b(x)=(log_k(x))/(log_k(b))`

So we have:

`1/(log_6 a)+1/(log_4 a)=1/((log_(24) a)/(log_(24) 6))+1/((log_(24)a)/(log_(24) 4))=(log_24 6)/(log_24 a)+(log_24 4)/(log_24 a)=(log_24 6+log_24 4)/(log_24 a)`

Now we use formula for logarithm of product:

`log_b(xy)=log_b x+log_b y`

So our expression is equal to:

`(log_24(6cdot4))/(log_24 a)=(log_24 24)/(log_24 a)=1/(log_24 a)`

For more on logarithms see the ling below.

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