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Find the value of
log_(5)
qquad (AHSME 1950)
What is the logarithm of
27root(4)(9)root(3)(9) base 3 ? (AHSME 1953)

$1$ . Find the value of $\log _{5}$ $\qquad$ (AHSME $1950$ ) $\newline$ $2$ . What is the logarithm of $27 \sqrt[4]{9} \sqrt[3]{9}$ base $3$ ? (AHSME $1953$ )

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Properties of logarithms: mixed review

Full solution

Q.

1

. Find the value of

\log _{5}

\qquad

(AHSME

1950

)

\newline

2

. What is the logarithm of

27 \sqrt[4]{9} \sqrt[3]{9}

base

3

? (AHSME

1953

)

Simplify Radicals: First, simplify $\sqrt[4]{27}$ and $\sqrt[3]{9}$ . $\newline$ - $\sqrt[4]{27} = 27^{1/4} = (3^3)^{1/4} = 3^{3/4}$ $\newline$ - $\sqrt[3]{9} = 9^{1/3} = (3^2)^{1/3} = 3^{2/3}$
Multiply Simplified Forms: Multiply the simplified forms. $\newline$ - $3^{3/4} \cdot 3^{2/3}$ $\newline$ - Use the property of exponents: $a^m \cdot a^n = a^{m+n}$ $\newline$ - $3^{3/4 + 2/3}$
Add Exponents: Add the exponents $3/4$ and $2/3$ . $\newline$ - Convert to a common denominator: $3/4 = 9/12$ and $2/3 = 8/12$ $\newline$ - $9/12 + 8/12 = 17/12$ $\newline$ - So, $3^{3/4} \cdot 3^{2/3} = 3^{17/12}$
Find Logarithm: Find the logarithm base $3$ of $3^{17/12}$ . $\newline$ - $\log_3(3^{17/12})$ $\newline$ - Use the logarithmic identity: $\log_b(a^n) = n \log_b(a)$ $\newline$ - $17/12 \log_3(3)$ $\newline$ - $\log_3(3) = 1$ $\newline$ - $17/12 \cdot 1 = 17/12$

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