Understanding the Logarithm Problem: Calculating log$_9$ 27 + log$_8$ 32

The problem asks us to find the sum of two logarithms with different bases: log$_9$ 27 and log$_8$ 32. To solve this, we need to evaluate each logarithm separately and then add the results.

Key Logarithm Properties Used

We will use the definition of a logarithm and the property that allows us to simplify logarithms where the base and the argument are powers of the same number:

• Definition: $\log_b a = x$ means $b^x = a$.
• Property: $\log_{b^m} a^n = \frac{n}{m}$.

Step-by-Step Calculation of log$_9$ 27

We need to find the value of $\log_9 27$. Let this value be $x$. According to the definition of a logarithm:

$\log_9 27 = x \implies 9^x = 27$

We can express both 9 and 27 as powers of 3:

$9 = 3^2$

$27 = 3^3$

Substitute these into the equation:

$(3^2)^x = 3^3$

$3^{2x} = 3^3$

Since the bases are equal, the exponents must be equal:

$2x = 3$

Solve for $x$:

$x = \frac{3}{2}$

So, $\log_9 27 = \frac{3}{2}$.

Alternatively, using the property $\log_{b^m} a^n = \frac{n}{m}$:

$\log_9 27 = \log_{3^2} 3^3 = \frac{3}{2}$.

Step-by-Step Calculation of log$_8$ 32

Next, we need to find the value of $\log_8 32$. Let this value be $y$. According to the definition of a logarithm:

$\log_8 32 = y \implies 8^y = 32$

We can express both 8 and 32 as powers of 2:

$8 = 2^3$

$32 = 2^5$

Substitute these into the equation:

$(2^3)^y = 2^5$

$2^{3y} = 2^5$

Since the bases are equal, the exponents must be equal:

$3y = 5$

Solve for $y$:

$y = \frac{5}{3}$

So, $\log_8 32 = \frac{5}{3}$.

Alternatively, using the property $\log_{b^m} a^n = \frac{n}{m}$:

$\log_8 32 = \log_{2^3} 2^5 = \frac{5}{3}$.

Calculating the Sum log$_9$ 27 + log$_8$ 32

Now we need to add the values we found for each logarithm:

Sum = $\log_9 27 + \log_8 32 = \frac{3}{2} + \frac{5}{3}$

To add these fractions, we need to find a common denominator. The least common multiple (LCM) of 2 and 3 is 6.

Convert each fraction to have a denominator of 6:

$\frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6}$

$\frac{5}{3} = \frac{5 \times 2}{3 \times 2} = \frac{10}{6}$

Now, add the fractions:

Sum = $\frac{9}{6} + \frac{10}{6} = \frac{9 + 10}{6} = \frac{19}{6}$

Therefore, the value of log$_9$ 27 + log$_8$ 32 is $\frac{19}{6}$.

Revision Table: Key Concepts in Logarithms

Additional Information: Solving Logarithm Problems

Solving logarithm problems often involves expressing the numbers in terms of common bases. Recognizing that numbers like 4, 8, 16, 32 are powers of 2, or 9, 27, 81 are powers of 3, is very helpful. The property $\log_{b^m} a^n = \frac{n}{m}$ is a direct result of the change of base formula or simply converting to exponential form. For instance, if $\log_{b^m} a^n = y$, then $(b^m)^y = a^n$, which means $b^{my} = b^{\log_b (a^n)}$. If $a^n$ can be written as $b^k$, then $b^{my} = b^k$, so $my=k$. We know $k = \log_b a^n = n \log_b a$. So $my = n \log_b a$. If $a$ is a power of $b$, say $a=b^p$, then $a^n=(b^p)^n=b^{pn}$, so $k=pn$. $my=pn$. $y = \frac{pn}{m}$. Since $p = \log_b a$, we have $y = \frac{n \log_b a}{m}$. However, if $a$ is a power of $b$, specifically $a=b^1$ or $a=b^p$, a simpler approach is $\log_{b^m} a^n = \frac{\log a^n}{\log b^m} = \frac{n \log a}{m \log b}$. If $a=b^p$, this becomes $\frac{n \log b^p}{m \log b} = \frac{np \log b}{m \log b} = \frac{np}{m}$. In our case, for $\log_{3^2} 3^3$, $b=3, m=2, a=3, n=3$. Since $a=b^1$, $p=1$. So $\frac{n}{m} = \frac{3}{2}$. For $\log_{2^3} 2^5$, $b=2, m=3, a=2, n=5$. Since $a=b^1$, $p=1$. So $\frac{n}{m} = \frac{5}{3}$. The property $\log_{b^m} a^n = \frac{n}{m}$ works directly when $a=b$. When $a$ is a power of $b$, say $a=b^k$, the property is $\log_{b^m} (b^k)^n = \log_{b^m} b^{kn} = \frac{kn}{m}$. For $\log_{3^2} 3^3$, $b=3, m=2, k=1$ (since $a=3=3^1$), $n=3$. Result is $\frac{1 \times 3}{2} = \frac{3}{2}$. For $\log_{2^3} 2^5$, $b=2, m=3, k=1$ (since $a=2=2^1$), $n=5$. Result is $\frac{1 \times 5}{3} = \frac{5}{3}$. The property $\log_{b^m} a^n = \frac{n}{m}$ is a specific case of $\log_{b^m} a^n = \frac{n}{m} \log_b a$. If $a=b$, $\log_b a = \log_b b = 1$, so it becomes $\frac{n}{m}$. In our problem, the argument ($27$ and $32$) is a power of the base's base ($3$ and $2$). So $\log_{3^2} 3^3$ uses $a=3, b=3, m=2, n=3$. $\log_{2^3} 2^5$ uses $a=2, b=2, m=3, n=5$. The property is correctly applied as $\frac{n}{m}$.