Change-of-Base Formula for Logarithms - Examples, Exercises and Solutions

To solve this problem, let's simplify the given expression $\frac{\log_85}{\log_89}$.

• Step 1: Recognize that both the numerator and denominator have the same base, 8.
• Step 2: The division property of logarithms states that $\frac{\log_b M}{\log_b N} = \log_N M$.
• Step 3: Apply the division rule to the given expression: $\frac{\log_8 5}{\log_8 9} = \log_9 5$.

Thus, after simplifying, we see that $\frac{\log_85}{\log_89} = \log_9 5$.

Hence, the correct answer is $\log_9 5$, which corresponds to the choice 1.

To solve the problem of evaluating $\log_7 4$, we will use the change-of-base formula for logarithms.

The change-of-base formula is:

• $\log_b a = \frac{\log_k a}{\log_k b}$, where $k$ can be any base, commonly chosen as 10 (common logs) or $e$ (natural logs).

We will choose natural logarithms ($\ln$) for simplicity, therefore:

$\log_7 4 = \frac{\ln 4}{\ln 7}$

By applying the change-of-base formula, we find that the logarithm $\log_7 4$ can be expressed as $\frac{\ln 4}{\ln 7}$.

Upon examining the provided choices, we identify that choice 2: $\frac{\ln 4}{\ln 7}$ matches our result.

Therefore, the solution to the problem is $\frac{\ln 4}{\ln 7}$.

To solve the given expression $\frac{\log_{4x}9}{\log_{4x}a}$ using the change-of-base formula, follow these steps:

• Step 1: Apply the change-of-base formula to both the numerator and the denominator expressions.
  This gives us: $\log_{4x}9 = \frac{\log_a 9}{\log_a (4x)}$ and $\log_{4x}a = \frac{\log_a a}{\log_a (4x)}$.
• Step 2: Substitute these into our original expression:
  $\frac{\log_{4x}9}{\log_{4x}a} = \frac{\frac{\log_a 9}{\log_a (4x)}}{\frac{\log_a a}{\log_a (4x)}}$.
• Step 3: Simplify the fraction:
  The $\log_a (4x)$ cancels out from the numerator and the denominator, leaving us with $\frac{\log_a 9}{\log_a a}$.
• Step 4: Further simplify using the fact that $\log_a a = 1$ because any number $a$ to the power of 1 is $a$.
  This results in $\frac{\log_a 9}{1} = \log_a 9$.

Therefore, the expression simplifies to $\log_a 9$.

The correct answer is $\log_a 9$, which matches choice 1.

To solve this problem, we'll simplify the given expression $\frac{\log_9e^2}{\log_9e}$ using logarithmic rules:

Step 1: Apply the power rule of logarithms:
The numerator $\log_9e^2$ can be rewritten using the power rule: $\log_9e^2 = 2 \cdot \log_9e$.

Step 2: Substitute and simplify the fraction:
Substitute the expression from Step 1 into the original problem:
$\frac{\log_9e^2}{\log_9e} = \frac{2 \cdot \log_9e}{\log_9e}$.

Step 3: Cancel common terms:
Since $\log_9e$ appears in both the numerator and the denominator, it cancels out, leaving:
$\frac{2 \cdot \log_9e}{\log_9e} = 2$.

Therefore, the solution to the problem is $\boxed{2}$.

To solve this problem, we'll follow these steps:

• Step 1: Apply the mathematical property using the quotient rule for logarithms.
• Step 2: Simplify the expression into the desired format.

Now, let's work through each step:
Step 1: Given the expression $\frac{\log_8 9a}{\log_8 3a}$, we can directly apply the quotient rule for logarithms, which tells us that $\frac{\log_b M}{\log_b N} = \log_N M$.
Step 2: Applying this formula, we find that $\frac{\log_8 9a}{\log_8 3a} = \log_{3a} 9a$.

Therefore, the solution to the problem is $\log_{3a} 9a$.

To solve this problem, we’ll follow these steps:

• Step 1: Identify the expression $\ln 4x$.
• Step 2: Apply the change-of-base formula to transform the natural logarithm to one using base 7.
• Step 3: Use the formula $\ln a = \frac{\log_b a}{\log_b e}$ to substitute the values.

Now, let's work through each step:
Step 1: The expression given is $\ln 4x$.
Step 2: We want a base 7 logarithm, so we apply the change-of-base formula:
Step 3: We have:

$\ln 4x = \frac{\log_7 4x}{\log_7 e}$

Therefore, the logarithmic expression $\ln 4x$ in base 7 is equivalent to $\frac{\log_7 4x}{\log_7 e}$.

This matches the correct answer choice, which is choice 4.

To solve the problem, we'll follow these steps:

• Step 1: Simplify the given expression using logarithmic identities.
• Step 2: Solve the resulting quadratic equation for $x$.

Now, let's work through each step:

Step 1: We start with the equation:

$\frac{\log_4(x^2+8x+1)}{\log_4 8} = 2$

We know that $\log_4 8 = \frac{3}{2}$, since $8 = 4^{3/2}$. Thus, we can rewrite the equation as:

$\log_4(x^2+8x+1) = 2 \times \frac{3}{2} = 3$

Applying the property of logarithms that states $\log_b a = c \Rightarrow a = b^c$, we have:

$x^2 + 8x + 1 = 4^3 = 64$

Step 2: Solve the resulting quadratic equation:

$x^2 + 8x + 1 = 64$

Subtract 64 from both sides to bring the equation to standard form:

$x^2 + 8x + 1 - 64 = 0$

$x^2 + 8x - 63 = 0$

Now, apply the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 8$, and $c = -63$:

$x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1}$

$x = \frac{-8 \pm \sqrt{64 + 252}}{2}$

$x = \frac{-8 \pm \sqrt{316}}{2}$

Simplify $\sqrt{316}$ as $\sqrt{79 \cdot 4} = 2\sqrt{79}$:

$x = \frac{-8 \pm 2\sqrt{79}}{2}$

Thus, $x = -4 \pm \sqrt{79}$.

Therefore, the solution to the equation is $x = -4 \pm \sqrt{79}$.

To solve the given equation $\frac{\log_8(4x) + \log_8(x+2)}{\log_8 3} = 3$, we follow these steps:

• Step 1: Combine the logs in the numerator using the product rule

  We use the product rule: $\log_8(4x) + \log_8(x+2) = \log_8((4x)(x+2)) = \log_8(4x^2 + 8x)$.

• Step 2: Equate the fraction to 3 and solve the resulting equation

  This gives us $\frac{\log_8(4x^2 + 8x)}{\log_8 3} = 3$.

  Cross-multiplying, we have $\log_8(4x^2 + 8x) = 3\log_8 3$.

  By the power rule, we can simplify as $\log_8(4x^2 + 8x) = \log_8 3^3 = \log_8 27$.

• Step 3: Solve for $x$

  Since the logarithms are the same base, we equate the arguments: $4x^2 + 8x = 27$.

  Rearranging gives the quadratic equation $4x^2 + 8x - 27 = 0$.

  We solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 4$, $b = 8$, and $c = -27$.

  Thus, $x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 4 \cdot (-27)}}{2 \cdot 4}$.

  Calculating further, $x = \frac{-8 \pm \sqrt{64 + 432}}{8}$.

  This simplifies to $x = \frac{-8 \pm \sqrt{496}}{8}$.

  Simplifying $\sqrt{496} = 4\sqrt{31}$, the equation becomes:

  $x = \frac{-8 \pm 4\sqrt{31}}{8}$.

  Further simplifying gives us two solutions: $x = -1 \pm \frac{\sqrt{31}}{2}$.

Given that $x$ must be positive for the original logarithms to be valid, we take $x = -1 + \frac{\sqrt{31}}{2}$.

Therefore, the correct solution is $x = -1+\frac{\sqrt{31}}{2}$.

To solve this problem, we will rewrite both sides of the inequality with the change of base formula and evaluate them:

• Step 1: Rewrite both logarithms using common base.
  Using the change of base formula, $\log_{\frac{1}{4}}9 = \frac{\log_29}{\log_2\frac{1}{4}}$ and $\frac{\log_57}{\log_5\frac{1}{4}}$ can also be rewritten similarly.
• Step 2: Recognize that changing everything to base 2 will be beneficial.
  $\log_2 (1/4) = \log_2 4^{-1} = -2$.
• Step 3: Evaluate left side.
  $\log_{\frac{1}{4}}9 = \frac{\log_2 9}{-2} = -\frac{\log_2 9}{2}$.
• Step 4: Evaluate right side using the same base.
  $\frac{\log_57}{\log_5\frac{1}{4}} = -\frac{\log_2 7}{2}$, where $-\log_2 4 = 2$.
• Step 5: Compare both expressions
  $-\frac{\log_2 9}{2} < -\frac{\log_2 7}{2} => \log_2 9 < \log_2 7 => 9 < 7$.
• Step 6: Since $9 > 7$, convert the logarithmic expressions back into $\log_{\frac{1}{4}}$.
  $\log_{\frac{1}{4}}7$ is smaller than $\log_{\frac{1}{4}}9$, so inequation holds.

After comparing these expressions, we see that $\log_{\frac{1}{4}}9 < \log_{\frac{1}{4}}7$ indeed holds true.

Therefore, the solution is: Yes, since: $\log_{\frac{1}{4}}9 < \log_{\frac{1}{4}}7$.

To solve this problem, we'll follow these steps:

• Step 1: Combine the logarithms in the numerator.
• Step 2: Simplify the expression using logarithmic properties.

Now, let's work through each step:

Step 1: Combine the logarithms in the numerator using the sum of logarithms property:

$\log_45 + \log_42 = \log_4(5 \times 2) = \log_4 10.$

Step 2: Simplify the entire expression $\frac{\log_4 10}{3\log_4 2}$:

$\frac{\log_4 10}{3 \log_4 2} = \frac{\log_4 10}{\log_4 2^3} = \frac{\log_4 10}{\log_4 8} = \log_8 10.$

This follows from the property that $\frac{\log_b x}{\log_b y} = \log_y x$.

Therefore, the solution to the problem is $\log_8 10$.

To solve the problem $\frac{2\log_7 8}{\log_7 4} + \frac{1}{\log_4 3} \times \log_2 9$, we will apply various logarithmic rules:

Step 1: Simplify $\frac{2\log_7 8}{\log_7 4}$.

• Using the power property, $\log_7 8 = \log_7 2^3 = 3\log_7 2$.
• Similarly, $\log_7 4 = \log_7 2^2 = 2\log_7 2$.
• The expression becomes $\frac{2 \times 3\log_7 2}{2\log_7 2} = 3$.

Step 2: Simplify $\frac{1}{\log_4 3} \times \log_2 9$.

• $\frac{1}{\log_4 3} = \log_3 4$, by inversion.
• $\log_2 9$ can be expressed as $\log_2 3^2 = 2\log_2 3$.
• The product becomes $\log_3 4 \times 2\log_2 3 = 2 \cdot \frac{\log_2 4}{\log_2 3} \times \log_2 3$.
• Since $\log_2 4 = 2$, this simplifies to $2 \times \frac{2}{1} = 4$.

Step 3: Add the results from Steps 1 and 2:
$3 + 4 = 7$.

Therefore, the solution to the problem is $7$.

To solve this problem, we'll proceed as follows:

• Step 1: Rewrite each logarithmic expression using the change of base formula.
• Step 2: Simplify the expressions using properties of logarithms.
• Step 3: Identify the final expression.

Now, let's work through each step:

Step 1: We begin by converting each logarithm to the natural logarithm base.
Using the change of base formula, we have:

$\frac{\log_3 11}{\log_3 4} = \frac{\frac{\ln 11}{\ln 3}}{\frac{\ln 4}{\ln 3}} = \frac{\ln 11}{\ln 4}$.

Step 2: Next, simplify the second expression:

$\frac{1}{\ln 3} \cdot 2\log 3 = 2$.

This follows because $\log 3$ in natural logarithms converts to $\ln 3$, and thus:

$\frac{2\ln 3}{\ln 3} = 2$.

Hence, our entire expression now is $\frac{\ln 11}{\ln 4} + 2$.

Step 3: Express $2$ as a logarithm. Using the properties of logarithms:

$2 = \log e^2$, since $\ln e = 1$.

Therefore, the entire expression becomes:

$\frac{\ln 11}{\ln 4} + \log e^2$.

By the properties of logarithms, this can also be expressed as:

$\log_4 11 + \log e^2$.

Thus, the expression simplifies directly to:

$\log_4 11 + \log e^2$.

Therefore, the solution to the problem is $\log_4 11 + \log e^2$.

To solve this problem, we'll simplify the expression step-by-step, using algebraic rules for logarithms:

• Step 1: Simplify the numerator $\frac{\log_7 6 - \log_7 1.5}{3 \log_7 2}$

First, apply the logarithm quotient rule to the numerator:
$\log_7 6 - \log_7 1.5 = \log_7 \left(\frac{6}{1.5}\right) = \log_7 4$

• Step 2: Simplify $3 \log_7 2$ in the denominator.

The denominator is $3 \times \log_7 2$.

• Step 3: Address the next part of the expression: $\frac{1}{\log_{\sqrt{8}} 2}$.

By changing the base, use $\log_{\sqrt{8}} 2 = \frac{\log_{8} 2}{\frac{1}{2}}$ because $\sqrt{8} = 8^{1/2}$. Now, $\log_8 2 = \frac{1}{3}$ as $8^{1/3} = 2$. So, $\log_{\sqrt{8}} 2 = \frac{\log_2 8}{1/2} = \frac{1/3}{1/2} = \frac{2}{3}$.

Therefore, the reciprocal is $\frac{1}{\log_{\sqrt{8}} 2} = \frac{3}{2}$.

• Step 4: Combine and simplify the expression.

The complete logarithmic expression simplifies as follows:
$\frac{\log_7 4}{3 \log_7 2} \cdot \frac{3}{2} = \frac{\log_7 (2^2)}{3 \log_7 2} \cdot \frac{3}{2}$

Using the power rule, $\log_7 4 = 2 \log_7 2$. Plug this back into the expression:
$\frac{2 \log_7 2}{3 \log_7 2} \cdot \frac{3}{2}$
The $\log_7 2$ cancels within the fraction, and we are left with $\frac{2}{3} \times \frac{3}{2} = 1$.

Therefore, the solution to the problem is $1$.

To solve this problem, we'll follow these steps:

• Step 1: Apply the change-of-base formula to $\frac{\ln 4}{\ln 5}$.

• Step 2: Apply the reciprocal property to $\frac{1}{\log_6 5}$.

• Step 3: Use the subtraction property of logs to simplify the expression.

• Step 4: Combine the simplified logarithms and multiply by -3.

Now, let's work through each step:

Step 1: Using the change-of-base formula, we have $\frac{\ln 4}{\ln 5} = \log_5 4$.

Step 2: Apply the reciprocal property to the third term: $\frac{1}{\log_6 5} = \log_5 6$.

Step 3: Substitute into the expression: $-3(\log_5 4 - \log_5 7 + \log_5 6)$.

Step 4: Combine terms using the properties of logs: $\log_5 4 - \log_5 7 + \log_5 6 = \log_5 \left(\frac{4 \times 6}{7}\right)$.

Step 5: Simplify to get: $\log_5 \left(\frac{24}{7}\right)$.

Multiply by -3: $-3(\log_5 (\frac{24}{7})) = 3\log_5 \left(\frac{7}{24}\right)$.

Therefore, the solution to the problem is $3\log_5 \frac{7}{24}$.

To solve the inequality $\frac{\log_{\frac{1}{8}}(2x)}{\log_{\frac{1}{8}}(4)} < \log_4(5x - 2)$, we proceed as follows:

• Step 1: Convert all logarithms to a common base using the change of base formula:

  $\log_{\frac{1}{8}}(a) = \frac{\log(a)}{\log(\frac{1}{8})}$ and $\log_4(b) = \frac{\log(b)}{\log(4)}$.

• Step 2: Simplify the inequality using these conversions.

  The left expression becomes $\frac{\log(2x)}{\log(\frac{1}{8})} \div \frac{\log(4)}{\log(\frac{1}{8})} = \frac{\log(2x)}{\log(4)}$.

• Step 3: The inequality simplifies to $\frac{\log(2x)}{\log(4)} < \frac{\log(5x - 2)}{\log(4)}$.
• Step 4: Since both sides are divided by the positive $\log(4)$, the inequality remains:

  $\log(2x) < \log(5x - 2)$.

• Step 5: Remove logs since the logarithms are to the same base, leading to $2x < 5x - 2$.
• Step 6: Solve the inequality $2x < 5x - 2$. Rearrange terms: $2 < 3x$.
• Step 7: Divide both sides by 3 to solve for $x$: $\frac{2}{3} < x$.
• Step 8: Validate $(5x - 2) > 0$ implies $x > \frac{2}{5}$, which is consistent with our solution.

Therefore, the solution to the problem is $\frac{2}{3} < x$, which is choice 1.