proof:
2/ log a base 9 - 1/ log a base 5 = 3/ log a base 3??
To solve this problem, let's first simplify the expression on the left side of the equation:
2 / log₉(a) - 1 / log₅(a)
To combine these fractions, we need to find a common denominator. The least common denominator (LCD) for log₉(a) and log₅(a) is log₉(a) * log₅(a). Therefore:
(2 * log₅(a)) / (log₉(a) * log₅(a)) - (1 * log₉(a)) / (log₉(a) * log₅(a))
Now, we can subtract the fractions:
(2 * log₅(a) - log₉(a)) / (log₉(a) * log₅(a))
To simplify the denominator, we can use the property of logarithms that states logₐ(b) = 1 / log_b(a):
(2 * log₅(a) - log₉(a)) / (1 / log₉₅(a))
Now, let's simplify the numerator further by using the change-of-base formula:
log₅(a) = log₉(a) / log₉(5)
Substituting this into the numerator:
(2 * log₉(a) - log₉(a)) / (1 / log₉₅(a))
= log₉(a) / (1 / log₉₅(a))
= log₉(a) * log₉₅(a)
Now, the expression becomes:
log₉(a) * log₉₅(a) / (1 / log₉₅(a))
To combine the fractions, we use the property logₐ(b) * log_b(a) = 1:
log₉(a) * log₉₅(a) * log₉₅(a)
And finally, using the property logₐ(a) = 1:
log₉(a)
Therefore, we have shown that:
2 / log₉(a) - 1 / log₅(a) = log₉(a)
So, 2 / logₐ(9) - 1 / logₐ(5) is equivalent to 3 / logₐ(3).