prove:
1/log(base36)a = 2/log(base6)a
i made a common base and then multiplied both sides with log(base6)a to get
2=log(base6)a/log(base6^2)a but this is as far as i can figure.
since 36=6^2
its exponents are only half as big as powers of 6 to get to the same value. That is,
log_36(n) = 1/2 log_6(n)
So,
1/log_36(a) = 1/ (1/2 log_6(a)) = 2/log_6(a)
To prove the equation 1/log(base36)a = 2/log(base6)a, we can use the change of base formula for logarithms. The change of base formula states that for any positive real numbers a, b, and c where a ≠ 1, log(basea)c = log(baseb)c / log(baseb)a.
Let's use this formula to rewrite the equation:
1/log(base36)a = 2/log(base6)a
Rewriting the logarithms using the change of base formula:
log(basea)b = log(basec)b / log(basec)a
1/(log(base36)a) = 2/(log(base6)a / log(base6)36)
Simplifying the expression on the right side:
1/(log(base36)a) = 2/(log(base6)a / log(base6)6^2)
Since log(base6)6^2 = 2 (log(base6)6 = 1), the equation becomes:
1/(log(base36)a) = 2/2
Simplifying further:
1/(log(base36)a) = 1
Since both sides are equal to 1, the equation is proved:
1/log(base36)a = 2/log(base6)a.