High School

Showing all steps in your calculation, simplify: log_a(a⁻⁰. ⁰⁰⁸0.008) +log_5 (0.008). 4. Find the first derivative of the function h(x) = e^exp(x)=x

Answer :

Final Answer:

The simplified expression is: log_a(0.2) + log_5(0.008)

Explanation:

To simplify the given expression, we'll start by using logarithmic properties. We'll use the fact that log_a(b) + log_a(c) = log_a(b * c) and log_a([tex]a^n[/tex]) = n for any base 'a' and real numbers 'b', 'c', and 'n'.

First, let's work with the logarithm log_a([tex]a^(^-^0^.^0^0^8^)[/tex]). Since any nonzero number raised to the power of -0.008 is equivalent to its reciprocal raised to the power of 0.008 (i.e., [tex]a^(^-^0^.^0^0^8^)[/tex]= 1/[tex]a^(^-^0^.^0^0^8^)[/tex]), we have log_a([tex]a^(^-^0^.^0^0^8^)[/tex]) = log_a(1/[tex]a^(^-^0^.^0^0^8^)[/tex]).

Now, using the property log_a(1/b) = -log_a(b) for any positive number 'b', we can rewrite this as -log_a[tex]a^(^-^0^.^0^0^8^)[/tex], which simplifies to -0.008 (since log_a([tex]a^n[/tex]) = n).

Next, we have log_5(0.008). This is the logarithm of 0.008 with base 5. We can rewrite 0.008 as [tex]5^(^-^3^)[/tex], so log_5(0.008) = log_5([tex]5^(^-^3^)[/tex]).

Using the logarithmic property log_b([tex]b^n[/tex]) = n, we get log_5([tex]5^(^-^3^)[/tex]) = -3.

Now, we can simplify the original expression by adding these two simplified logarithmic terms: log_a([tex]a^(^-^0^.^0^0^8^)[/tex]) + log_5(0.008) = -0.008 + (-3) = -3.008.

Thus, the final simplified expression is -3.008.

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