The value of $\log_{0.008} \sqrt{5} + \log_{\sqrt{3}} 81 - 7$ is equal to

We need to evaluate $\log_{0.008} \sqrt{5} + \log_{\sqrt{3}} 81 - 7$.

Let us break this down into manageable pieces and solve each logarithm step by step.

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Evaluating $\log_{0.008} \sqrt{5}$

First, we'll rewrite both the base and the argument in more convenient forms:

$0.008 = \frac{8}{1000} = \frac{2^3}{10^3} = \left(\frac{2}{10}\right)^3 = (0.2)^3$

$\sqrt{5} = 5^{1/2}$

So our expression becomes: $\log_{(0.2)^3} 5^{1/2}$

Using the logarithm property $\log_{a^m} b^n = \frac{n}{m} \log_a b$:

$\log_{(0.2)^3} 5^{1/2} = \frac{1/2}{3} \log_{0.2} 5 = \frac{1}{6} \log_{0.2} 5$

Now we need to find $\log_{0.2} 5$:

Since $0.2 = \frac{1}{5}$, we have $\log_{0.2} 5 = \log_{1/5} 5$

Using the property $\log_{1/a} b = -\log_a b$: $\log_{1/5} 5 = -\log_5 5 = -1$

Therefore: $\log_{0.008} \sqrt{5} = \frac{1}{6} \times (-1) = -\frac{1}{6}$

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Evaluating $\log_{\sqrt{3}} 81$

Let us rewrite this using exponential forms:

$81 = 3^4$

$\sqrt{3} = 3^{1/2}$

So our expression becomes: $\log_{3^{1/2}} 3^4$

Using the logarithm property $\log_{a^m} a^n = \frac{n}{m}$:

$\log_{3^{1/2}} 3^4 = \frac{4}{1/2} = 4 \times 2 = 8$

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Now we can substitute back into the original expression:

$\log_{0.008} \sqrt{5} + \log_{\sqrt{3}} 81 - 7$

$= -\frac{1}{6} + 8 - 7$

$= -\frac{1}{6} + 1$

$= -\frac{1}{6} + \frac{6}{6}$

$= \frac{5}{6}$

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When dealing with logarithms with unusual bases, always try to express both the base and argument as powers of the same number. This allows us to use the property $\log_{a^m} a^n = \frac{n}{m}$, which makes calculations much simpler.

Answer: $\frac{5}{6}$