Prove that A is a constant A = (log_5_x^2 + log_5_x^3)/(log_4_x^2 + log_4_x^3)

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neela | High School Teacher | (Level 3) Valedictorian

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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.


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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

We'll simplify:

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)

We'll simplify:

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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