Hello,

Is log_3 (5) equal to log_5 (3)? Explain your answer. Do not evaluate the logarithms.

is my question. What is a surefire way to show this is false.

let x = log (base3) 5

then 3^x = 5 , clearly x > 1 ,
because 3^1 = 3 and 3^2 = 25 ,so x must be between 1 and 2

let y = log(base5)^3
then 5^y = 3 , clearly y < 1,
because 5^0 = 1 and 5^1 = 5
so y has to be between 0 and 1
so x ≠ y
and the two log expressions cannot be equal

fantastic, makes so much sense now, thx

and btw 3^2 = 25 is typo

To determine if log_3(5) is equal to log_5(3), we can use a property of logarithms called the change of base formula. This formula allows us to rewrite a logarithm with any base in terms of logarithms with a different base.

The change of base formula states that for any positive numbers a, b, and c:

log_a(c) = log_b(c) / log_b(a)

In this case, we want to compare log_3(5) and log_5(3), so we can rewrite both logarithms using the same base, such as the natural logarithm (log_e):

log_3(5) = log_e(5) / log_e(3)
log_5(3) = log_e(3) / log_e(5)

Now, if log_3(5) is equal to log_5(3), the fractions on the right side of these equations should be equal as well.

So, to show that log_3(5) is not equal to log_5(3), we need to demonstrate that the fractions log_e(5) / log_e(3) and log_e(3) / log_e(5) are not equal.

One way to do this is by taking the ratio of these fractions:

(log_e(5) / log_e(3)) / (log_e(3) / log_e(5))

We can simplify this equation further by multiplying the numerator and denominator by the denominators of the fractions:

(log_e(5) / log_e(3)) * (log_e(5) / log_e(5)) / (log_e(3) / log_e(5)) * (log_e(3) / log_e(3))

This simplifies to:

(log_e(5) * log_e(5)) / (log_e(3) * log_e(3))

Now, if log_3(5) and log_5(3) were equal, this ratio should be equal to 1. However, if you evaluate this expression, you will find that it is not equal to 1. Therefore, log_3(5) is not equal to log_5(3).

In summary, to prove that log_3(5) is not equal to log_5(3), we used the change of base formula to rewrite these logarithms with the same base. Then, we compared the resulting fractions and found that they are not equal, disproving the equality between the two logarithms.