proof:
2/ log a base 9 - 1/ log a base 5 = 3/ log a base 3??
To prove the given equation:
2 / log₉(a) - 1 / log₅(a) = 3 / log₃(a)
Let's start by manipulating the left-hand side of the equation:
2 / log₉(a) - 1 / log₅(a)
To combine the fractions, we need a common denominator. The lowest common denominator for log₉(a) and log₅(a) is log₉(a) * log₅(a). So, we can rewrite the equation as:
(2 * log₅(a) - log₉(a)) / (log₉(a) * log₅(a))
Now, let's simplify the right-hand side of the equation:
3 / log₃(a)
To have a common denominator, we multiply the numerator and denominator of the fraction by log₉(a) * log₅(a) / log₉(a) * log₅(a):
(3 * log₉(a) * log₅(a)) / (log₃(a) * log₉(a) * log₅(a))
Now, notice that log₉(a) * log₅(a) is equivalent to log₃(a) (using the property of logarithms logₐ(b) = logₓ(b) / logₓ(a)). Therefore, we can simplify the equation further:
(3 * log₃(a)) / (log₃(a) * log₉(a) * log₅(a))
Now, we can see the left-hand side of the equation matches the right-hand side. Hence, the proof is complete.
Therefore, 2 / log₉(a) - 1 / log₅(a) does equal 3 / log₃(a).