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Using the logarithmic identities, you may convert the base of logarithm into a new base, such that:
`log_a b = (ln b)/(ln a)`
Reasoning by analogy yields:
`log_5 x^2 = (ln x^2)/(ln 5)`
`log_5 x^3 = (ln x^3)/(ln 5)`
`log_4 x^2 = (ln x^2)/(ln 4)`
`log_4 x^3 = (ln x^3)/(ln 4)`
Replacing the converted logarithms in fraction yields:
`A = ((ln x^2)/(ln 5) + (ln x^3)/(ln 5))/((ln x^2)/(ln 4) + (ln x^3)/(ln 4))`
`A = ((ln x^2 + ln x^3)/(ln 5))/((ln x^2 + ln x^3)/(ln 4))`
Reducing duplicate factors yields:
`A = (1/(ln 5))/(1/(ln 4)) => A = ln 4/ln 5 => A = log_5 4`
Hence, evaluating the given expression yields `A = log_5 4` and it does not contain x.
We notice that the numerator is a sum of logarithms that have matching bases.
We'll use the rule of product:
log a + log b = log (a*b)
log_5_x^2 + log_5_x^3 = log_5_(x^2*x^3)
log_5_(x^2*x^3) = log_5_x^(2+3)
log_5_x^2 + log_5_x^3 = log_5_x^5
We'll use the power rule of logarithms:
log_5_x^5 = 5*log_5_x (1)
We also notice that the denominator is a sum of logarithms that have matching bases.
log_4_x^2 + log_4_x^3 = log_4_x^5
log_4_x^2 + log_4_x^3 = 5*log_4_x (2)
We'll substitute both numerator and denominator by (1) and (2):
A = 5*log_5_x/5*log_4_x
A = log_5_x/log_4_x
We'll transform the base of the numerator, namely 5, into the base 4.
log_4_x = (log_5_x)*(log_4_5)
We'll re-write A:
A = log_5_x/(log_5_x)*(log_4_5)
A = 1/log_4_5
A = log_5_4
As we can notice, the result is a constant and it's not depending on the variable x.
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