Exploring Logarithmic Expressions | Understanding log₅(1/25) and Its Properties

log₅(1/25)

To solve the logarithmic expression log₅(1/25), we need to use the properties of logarithms

To solve the logarithmic expression log₅(1/25), we need to use the properties of logarithms. The logarithm of a number y to the base a is the exponent to which we must raise a to get y.

In this case, we want to find the exponent to which we must raise 5 to obtain the value 1/25.

Rewriting 1/25 as a power of 5, we have 1/25 = (1/5)², where (1/5) is the reciprocal of 5.

Now, we can rewrite the logarithmic expression as log₅((1/5)²), using the property logₐ(b^c) = c * logₐ(b).

Distributing the power of 2 inside the parentheses, we have log₅(1/5)².

Next, we use another property of logarithms, logₐ(b^c) = c * logₐ(b), to bring down the exponent 2 as a coefficient:
2 * log₅(1/5).

Now, we can simplify further. Since 1/5 = 5^(-1), we rewrite the expression as:
2 * log₅(5^(-1)).

Using the property logₐ(b^c) = c * logₐ(b) again, we can bring down the exponent -1:
2 * (-1) * log₅(5).

Simplifying, we have:
-2 * log₅(5).

At this point, we can evaluate the logarithm using a calculator or further simplify if necessary.

So, log₅(1/25) is equal to -2 * log₅(5).

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