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Chapter 13: Problem 32

Write as the sum or difference of logarithms and simplify, if possible. Assume all variables represent positive real numbers. $$\log \sqrt[3]{10}$$

Short Answer

Expert verified
The simplified expression for \(\log \sqrt[3]{10}\) is \(\boxed{\frac{1}{3}}\).

Step by step solution

01

Apply the cube root to the logarithm

By using the power rule for logarithms, we can rewrite the cube root as a fraction exponent within the logarithm. The power rule states that \(\log_{a}(b^{c}) = c \log_{a}(b)\). In this case, we have: \(\log \sqrt[3]{10} = \log(10^{\frac{1}{3}})\)
02

Use the power rule to rewrite the logarithm

Now that we have the logarithm expressed with a fractional exponent, we can apply the power rule to rewrite the expression as a product: \(\log(10^{\frac{1}{3}}) = \frac{1}{3}\log(10)\)
03

Simplify the expression

Since 10 is a base-10 logarithm, we know that \(\log(10) = 1\). Therefore, we can simplify the expression further: \(\frac{1}{3}\log(10) = \frac{1}{3}(1) = \frac{1}{3}\) Now we have expressed the given logarithm as the sum or difference of simpler logarithms and simplified the expression. The final answer is \(\boxed{\frac{1}{3}}\).

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Most popular questions from this chapter

If \(f(x)=x^{3}-1,\) show that \(f^{-1}(x)=\sqrt[3]{x+1}\)

Given that \(\log 5=0.6990\) and \(\log 9=0.9542,\) use the propertics of logarithms to approximate the following $$\log 25$$

Use the formula \(A=P e^{r t}\). If \(\$ 2000\) is invested at \(6 \%\) interest compounded continuously, how long would it take a) for the investment to grow to \(\$ 2500 ?\) b) for the initial investment to double?

Given that \(\log 5=0.6990\) and \(\log 9=0.9542,\) use the propertics of logarithms to approximate the following $$\log \frac{1}{9}$$

If \(f(x)=\frac{3}{2} x-9,\) show that \(f^{-1}(x)=\frac{2}{3} x+6\)
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