Prove that the result of the ratio does not depend on x. A = (log_5_x^2 + log_5_x^3)/(log_4_x^2 + log_4_x^3)

Asked on by else9393

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.