Problem Statement: We need to find the value of the logarithm of 64 with base 8. The logarithm log_(b)a answers the question: ＂To what power must the base b be raised, to produce the number a?＂ In this case, we are looking for the power to which 8 must be raised to get 64.Understanding Logarithms: We know that 8 is 2 raised to the power of 3, i.e., 8 = 2^3. Similarly, 64 is 2 raised to the power of 6, i.e., 64 = 2^6. We can use these equalities to rewrite the logarithm in terms of base 2.Using Change of Base Formula: Using the change of base formula for logarithms, we can express log_(8)64 as log_(2^3)(2^6). This can be simplified by using the property of logarithms that log_(b^m)(a^n) = (n/m) * log_(b)a. Since a is the same as b in this case, log_(b)b = 1.Applying Logarithm Property: Applying the property, we get log_(2^3)(2^6) = (6/3) * log_(2)2. Since log_(2)2 = 1 (because 2 to the power of 1 is 2), we simplify this to (6/3) * 1.Calculating the Final Result: Calculating (6/3) * 1 gives us 2 * 1, which equals 2. Therefore, log_(8)64 = 2.

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Grade 8

Relationship between squares and square roots

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\log _{8} 64=

Problem Statement: We need to find the value of the logarithm of $64$ with base $8$ . The logarithm $\log_{b}a$ answers the question: ＂To what power must the base $b$ be raised, to produce the number $a$ ?＂ In this case, we are looking for the power to which $8$ must be raised to get $64$ .
Understanding Logarithms: We know that $8$ is $2$ raised to the power of $3$ , i.e., $8 = 2^3$ . Similarly, $64$ is $2$ raised to the power of $6$ , i.e., $64 = 2^6$ . We can use these equalities to rewrite the logarithm in terms of base $2$ .
Using Change of Base Formula: Using the change of base formula for logarithms, we can express $\log_{8}64$ as $\log_{2^3}(2^6)$ . This can be simplified by using the property of logarithms that $\log_{b^m}(a^n) = \frac{n}{m} \cdot \log_{b}a$ . Since $a$ is the same as $b$ in this case, $\log_{b}b = 1$ .
Applying Logarithm Property: Applying the property, we get $\log_{2^3}(2^6) = \left(\frac{6}{3}\right) \cdot \log_{2}2$ . Since $\log_{2}2 = 1$ (because $2$ to the power of $1$ is $2$ ), we simplify this to $\left(\frac{6}{3}\right) \cdot 1$ .
Calculating the Final Result: Calculating $(\frac{6}{3}) \times 1$ gives us $2 \times 1$ , which equals $2$ . Therefore, $\log_{8}64 = 2$ .