Problem 71
Question
Write logarithm as the sum and/or difference of logarithms of a single quantity. Then simplify, if possible. \(\log _{5}\left(\frac{1}{t^{3}}\right)^{d}\)
Step-by-Step Solution
Verified Answer
\(-3d \cdot \log_5(t)\)
1Step 1: Apply the Power Rule for Logarithms
The power rule for logarithms states that \( \log_b(a^n) = n \cdot \log_b(a) \). Apply this rule to bring down the exponent \(d\) from \(\left( \frac{1}{t^3} \right)^d\). This gives us: \[ \log_5 \left( \left( \frac{1}{t^3} \right)^d \right) = d \cdot \log_5 \left( \frac{1}{t^3} \right) \]
2Step 2: Apply the Quotient Rule for Logarithms
The quotient rule for logarithms is \( \log_b\left(\frac{a}{b}\right) = \log_b(a) - \log_b(b) \). Apply this rule to \( \log_5 \left( \frac{1}{t^3} \right) \). Since \( 1 \) can be written as \( t^0 \), we have:\[ \log_5 \left( \frac{1}{t^3} \right) = \log_5(1) - \log_5(t^3) \]
3Step 3: Simplify Using Known Logarithm Values and the Power Rule
Note that \( \log_5(1) = 0 \) because any log base of 1 is 0. Also, apply the power rule \( \log_b(a^n) = n \cdot \log_b(a) \) on \( \log_5(t^3) \) to get \( 3 \cdot \log_5(t) \). Therefore:\[ \log_5 \left( \frac{1}{t^3} \right) = 0 - 3 \cdot \log_5(t) \] Thus, simplifying gives:\[ -3 \cdot \log_5(t) \]
4Step 4: Combine Results
Substitute the simplified expression back into the equation from Step 1:\[ d \cdot \log_5 \left( \frac{1}{t^3} \right) = d \cdot (-3 \cdot \log_5(t)) \] Thus, the expression becomes:\[ -3d \cdot \log_5(t) \]
Key Concepts
Power Rule for LogarithmsQuotient Rule for LogarithmsLogarithm Simplification
Power Rule for Logarithms
In the realm of logarithms, the power rule is an essential tool that greatly simplifies expressions. It states that for any logarithm of the form \( \log_b(a^n) \), you can move the exponent \( n \) to the front, simplifying the expression to \( n \cdot \log_b(a) \). By harnessing this rule, we can bring down exponents and work with more straightforward logarithmic expressions.
In our example, we start with \( \log_5 \left( \left( \frac{1}{t^3} \right)^d \right) \). Here, the exponent \( d \) applies to the entire fraction. By using the power rule, we rewrite it as \( d \cdot \log_5 \left( \frac{1}{t^3} \right) \). This transformation is the first crucial step in breaking down complex logarithmic expressions into more manageable forms.
In our example, we start with \( \log_5 \left( \left( \frac{1}{t^3} \right)^d \right) \). Here, the exponent \( d \) applies to the entire fraction. By using the power rule, we rewrite it as \( d \cdot \log_5 \left( \frac{1}{t^3} \right) \). This transformation is the first crucial step in breaking down complex logarithmic expressions into more manageable forms.
Quotient Rule for Logarithms
The quotient rule for logarithms is another powerful simplification technique. It is used when you have a division inside a logarithm. Essentially, this rule tells us that \( \log_b\left(\frac{a}{b}\right) \) can be separated into \( \log_b(a) - \log_b(b) \).
Let's apply this rule to the expression from the previous step: \( \log_5 \left( \frac{1}{t^3} \right) \). First, observe that \( \frac{1}{t^3} \) is another way of writing \( t^0 \div t^3 \). We can thus express the logarithm as \( \log_5(1) - \log_5(t^3) \).
This separation is vital as it not only breaks down the expression but sets the stage for further simplification. Remember that \( \log_5(1) = 0 \) since any number to the 0th power is always 1, and the logarithm of 1 is 0 for any base. Now we're left with the simpler \( -\log_5(t^3) \).
Let's apply this rule to the expression from the previous step: \( \log_5 \left( \frac{1}{t^3} \right) \). First, observe that \( \frac{1}{t^3} \) is another way of writing \( t^0 \div t^3 \). We can thus express the logarithm as \( \log_5(1) - \log_5(t^3) \).
This separation is vital as it not only breaks down the expression but sets the stage for further simplification. Remember that \( \log_5(1) = 0 \) since any number to the 0th power is always 1, and the logarithm of 1 is 0 for any base. Now we're left with the simpler \( -\log_5(t^3) \).
Logarithm Simplification
Simplification of logarithms is about making expressions as concise as possible by applying known values and rules. After using both the power and quotient rules, we have \( -\log_5(t^3) \). Here, we revisit the power rule for a deeper simplification. We know \( \log_b(a^n) = n \cdot \log_b(a) \), so \( \log_5(t^3) \) simplifies to \( 3 \cdot \log_5(t) \).
This turns our expression into \( -3 \cdot \log_5(t) \). Simplification is not just about applying rules but about understanding these transformations that lead to cleaner, more manageable expressions.
Finally, integrating this simplified expression back into our equation with \( d \) gives us the refined form: \( -3d \cdot \log_5(t) \). Through these systematic steps, you reduce a complex logarithmic expression to a simple, elegant solution. This clarity not only makes calculations more straightforward but helps in a deeper understanding of logarithmic properties.
This turns our expression into \( -3 \cdot \log_5(t) \). Simplification is not just about applying rules but about understanding these transformations that lead to cleaner, more manageable expressions.
Finally, integrating this simplified expression back into our equation with \( d \) gives us the refined form: \( -3d \cdot \log_5(t) \). Through these systematic steps, you reduce a complex logarithmic expression to a simple, elegant solution. This clarity not only makes calculations more straightforward but helps in a deeper understanding of logarithmic properties.
Other exercises in this chapter
Problem 71
Solve for \(x\). See Example 3 . $$ \log _{8} x=0 $$
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In your own words, what is a one-to-one function?
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Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log \frac{2-5 x}{2(x+8)}=0 $$
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Let \(f(x)=\frac{1}{x}\) and \(g(x)=\frac{1}{x^{2}} .\) Find each of the following. $$ (f \circ g)(5 x) $$
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