To show A= (log5 (x^2)+log5 (x^3)/log4 (x^2+log4 x^3) is a constant.
log a (x) = log (x)/log a.
To show that A is a constant.
We know that log a (x) +loga (y) = loga(xy)
Also we can change the base b: log a(x)= loga/loga.
Therefore numerator = log5 (x^2+log5 (x^3) = log5 (x^2*x^3) = log5 (x^5) = logx^5/log5......(2).
Denominator : log4 (x^2)+log4 (x^3) = log4 (x^2*x^3) = log4 (x^5) = log x^5 /log4.......... (3).
We substitute the numerator by log x^5/log5 from (1) and denominator by logx^5/log4 in the given expression and we get:
A= (log5 (x^2)+log5 (x^3)/log4 (x^2+log4 x^3) =( logx^5/log5)/(logx^5/log4) = log4/log5.
A= log5 (x^2)+log5 (x^3)/log4 (x^2+log4 x^3) = log4/log5.
Therefore, A = log4/log5 is a constant.
To prove that A is a constant means that to prove that the result of the ratio does not depend on 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.