Log₃ [tex]\frac{1}{9}[/tex] + log₉ [tex]\frac{1}{9}[/tex] - [tex]\frac{1}{3}[/tex] log₂₇ [tex]\frac{1}{9}[/tex]

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вопрос

First, let's simplify each logarithm expression using the properties of logarithms:

log₃ [tex]\frac{1}{9}[/tex] can be rewritten as log₃ 3^(-2) since 1/9 is equivalent to 3^(-2). Using the property logₐ (a^b) = b*logₐ(a), this simplifies to -2.

log₉ [tex]\frac{1}{9}[/tex] can be rewritten as log₉ 3^(-2) since 1/9 is equivalent to 3^(-2). Using the property logₐ (a^b) = b*logₐ(a), this simplifies to -2/2 = -1.

For log₂₇ [tex]\frac{1}{9}[/tex], 27 can be expressed as 3^3. Rewriting the logarithm, log₂(3^(-2)), using the change of base formula logₐb = logₐ(c)/logₐ(b), we get -2/log₂(3).

Now, substituting these simplified expressions back into the original equation:

-2 + (-1) - [tex]\frac{1}{3}[/tex] * (-2/log₂(3))
= -2 - 1 + 2/3 log₂(3)
= -3 + 2/3 log₂(3)

Therefore, the simplified expression is -3 + 2/3 log₂(3).