Subtraction of Logarithms - Examples, Exercises and Solutions

To solve the problem, let's use the rules of logarithms:

• Step 1: Recognize that we are dealing with the subtraction of logarithms sharing the same base, which calls for the identity $\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)$.
• Step 2: Apply this identity to the expression $\log_7 5 - \log_7 2$.
• Step 3: Realize that this can thus be expressed as a single logarithm: $\log_7 \left(\frac{5}{2}\right)$.
• Step 4: Simplify the fraction, yielding $\log_7 2.5$.

Therefore, the simplification results in the expression: $\log_7 2.5$.

This matches the correct answer from the given choices.

To solve the problem, we employ the property of logarithms for subtraction:

• Step 1: Recognize the expression $\log_5 3 - \log_5 2$.
• Step 2: Apply the logarithmic property for subtraction, $\log_b a - \log_b c = \log_b \left( \frac{a}{c} \right)$.
• Step 3: Substitute into the property: $\log_5 3 - \log_5 2 = \log_5 \left( \frac{3}{2} \right)$.

By applying the property, we simplify the expression to $\log_5 \frac{3}{2}$. This is equivalent to $\log_5 1.5$. Therefore:

Therefore, the result of the expression is $\log_5 1.5$.

To solve the problem of evaluating $\log_2 9 - \log_2 3$, we apply the properties of logarithms as follows:

• Step 1: Recognize that the expression uses a subtraction of logarithms with the same base: $\log_2 9 - \log_2 3$.
• Step 2: Use the logarithmic subtraction rule: $\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)$.
• Step 3: Simplify using this rule: $\log_2 9 - \log_2 3 = \log_2 \left(\frac{9}{3}\right)$.
• Step 4: Perform the division: $\frac{9}{3} = 3$.
• Step 5: Therefore, $\log_2 \left(\frac{9}{3}\right) = \log_2 3$.

Thus, the simplified and evaluated result is $\log_2 3$.

To solve the problem $\frac{1}{2}\log_39-\log_31.5$, we need to apply the rules of logarithms:

• **Step 1: Simplify with the power rule**
  Using the power rule $\frac{1}{2}\log_39 = \log_39^{1/2}$. Since $9 = 3^2$, we have $9^{1/2} = 3^{2 \cdot 1/2} = 3^1$. Thus, $\frac{1}{2}\log_39 = \log_3 3 = 1$.
• **Step 2: Apply the subtraction rule**
  Now, the expression becomes $1 - \log_3 1.5$. Using the subtraction rule: $1 - \log_3 1.5 = \log_3 3 - \log_3 1.5 = \log_3 \left(\frac{3}{1.5}\right)$.
• **Step 3: Simplify the fraction**
  Calculate $\frac{3}{1.5}$: it simplifies to 2 because $3 \div 1.5 = 2$.

Thus, the simplified expression is $\log_3 2$.

Using the provided answer choices, the correct answer matches choice $\log_3 2$, which corresponds to choice 2.

Therefore, the solution to the problem is $\log_3 2$.

We break it down into parts

$\log_61296=x$

$6^x=1296$

$x=4$

$\frac{1}{4}\cdot4\cdot\log_6\frac{1}{2}-\log_63=$

$\log_6\frac{1}{2}-\log_63=$

$\log_6\left(\frac{1}{2}:3\right)=\log_6\frac{1}{6}$

$\log_6\frac{1}{6}=x$

$6^x=\frac{1}{6}$

$x=-1$

$\log_ax-\log_ay=\log_a\frac{x}{y}$

$\log_7x^4-\log_72x^2=$

$\log_7\frac{x^4}{2x^2}=3$

$7^3=\frac{x^2}{2}$

We multiply by: $2$

$2\cdot7^3=x^2$

Extract the root

$x=\sqrt{680}=7\sqrt{14}$

$x=-\sqrt{680}=-7\sqrt{14}$

Let's solve the logarithmic equation step-by-step:

Step 1: Combine the Logarithms
Using the property $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$, we combine the logarithms:

$\ln\left(\frac{4x+3}{x^2-8}\right) = 2$

Step 2: Remove the Logarithm by Exponentiation
Exponentiate both sides with base $e$ to get rid of the natural logarithm:

$\frac{4x+3}{x^2-8} = e^2$

Step 3: Solve the Resulting Equation
Multiplying both sides by $x^2 - 8$ to eliminate the fraction:

$4x + 3 = e^2(x^2 - 8)$

Expanding and rearranging gives us:

$e^2x^2 - 4x - 8e^2 - 3 = 0$

Let's employ the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = e^2$, $b = -4$, and $c = -(8e^2 + 3)$.

Calculate the discriminant:

$b^2 - 4ac = (-4)^2 - 4(e^2)(-(8e^2 + 3))$

Solving this using numerical approximations (since we have $e^2 \approx 7.39$), you get:

$x \approx 3.18$

Conclusion:
The value of $x$ is approximately $3.18$, which confirms our choice.

To solve this problem, we'll apply the following steps:

• Step 1: Use logarithmic properties to rewrite $\log_4(3x^2+8x-10) - \log_4(-x^2-x+12.5) = 0$ as a single logarithm: $\log_4 \left( \frac{3x^2 + 8x - 10}{-x^2 - x + 12.5} \right) = 0$.
• Step 2: Recognize that if $\log_4 \left( \frac{3x^2 + 8x - 10}{-x^2 - x + 12.5} \right) = 0$, then $\frac{3x^2 + 8x - 10}{-x^2 - x + 12.5} = 4^0 = 1$.
• Step 3: Set up the equation: $3x^2 + 8x - 10 = -x^2 - x + 12.5$.
• Step 4: Rearrange the equation to: $3x^2 + 8x - 10 + x^2 + x - 12.5 = 0$.
• Step 5: Simplify to: $4x^2 + 9x - 22.5 = 0$.
• Step 6: Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4$, $b = 9$, $c = -22.5$.
• Step 7: Calculate: $b^2 - 4ac = 9^2 - 4 \times 4 \times (-22.5) = 81 + 360 = 441$.
• Step 8: Find $x$: $x = \frac{-9 \pm \sqrt{441}}{8} = \frac{-9 \pm 21}{8}$.
• Step 9: Compute the roots: $x_1 = \frac{-9 + 21}{8} = \frac{12}{8} = 1.5$ and $x_2 = \frac{-9 - 21}{8} = \frac{-30}{8} = -3.75$.
• Step 10: Verify these solutions satisfy the domain of original logarithmic expressions by substituting back into $3x^2 + 8x - 10 > 0$ and $-x^2 - x + 12.5 > 0$.

Therefore, the solutions to the problem are $x = -3.75, 1.5$.

The correct choice from the provided options is: $-3.75, 1.5$.

To solve the equation $\log 4x + \log 2 - \log 9 = \log_2 4$, we will follow these steps:

• Step 1: Simplify the left side using logarithmic properties
• Step 2: Convert the right side using change of base
• Step 3: Equate the simplified expressions and solve for $x$

Step 1: Simplify the left side:

The left side $\log 4x + \log 2 - \log 9$ can be combined using the properties of logarithms:

$\log 4x + \log 2 = \log(4x \cdot 2) = \log(8x)$

Now, using the subtraction property:

$\log (8x) - \log 9 = \log \left(\frac{8x}{9}\right)$

Step 2: Convert the right side using the change of base formula:

$\log_2 4 = \frac{\log 4}{\log 2}$

We recognize that $4 = 2^2$, so $\log_2 4 = 2$.

Step 3: Equate the expressions and solve for $x$:

Now equate:

$\log \left(\frac{8x}{9}\right) = 2$

This implies:

$\frac{8x}{9} = 10^2 = 100$

Thus, solving for $x$:

$8x = 900$

$x = \frac{900}{8} = 112.5$

Therefore, the solution to the problem is $x = 112.5$.

We will solve the problem step by step:

Step 1: Simplify $\log_9 e^3$

• Using the change of base formula, $\log_9 e^3 = \frac{\ln e^3}{\ln 9}$.
• We know $\ln e^3 = 3\ln e = 3$, because $\ln e = 1$.
• Thus, $\log_9 e^3 = \frac{3}{\ln 9} = \frac{3}{2\ln 3}$, since $\ln 9 = 2\ln 3$.
• Therefore, $\log_9 e^3 = \frac{3}{2\ln 3}$.

Step 2: Simplify $\log_2 24 - \log_2 8$

• Use the logarithm subtraction rule: $\log_2 24 - \log_2 8 = \log_2 \left(\frac{24}{8}\right) = \log_2 3$.

Step 3: Simplify $\ln 8 + \ln 2$

• Using the product property of logarithms: $\ln 8 + \ln 2 = \ln(8 \times 2) = \ln 16$.
• Since $16 = 2^4$, $\ln 16 = 4\ln 2$.

Step 4: Combine the results

• We need to check the overall structure: $\log_9 e^3 \times \log_2 3 \times 4 \ln 2$.
• Previously calculated: $\log_9 e^3 = \frac{3}{2 \ln 3}$, $\log_2 3 = \frac{\ln 3}{\ln 2}$.
• Therefore, the entire expression becomes:
• $\frac{3}{2 \ln 3} \times \frac{\ln 3}{\ln 2} \times 4 \ln 2 = \frac{3}{2} \times 4 = 6$.

Therefore, the solution to the problem is $6$.

Defined domain

x>0

x+1>0

x>-1

$\log7x+\log\left(x+1\right)-\log7=\log2x-\log x$

$\log\frac{7x\cdot\left(x+1\right)}{7}=\log\frac{2x}{x}$

We reduce by: $7$ and by $X$

$x\left(x+1\right)=2$

$x^2+x-2=0$

$\left(x+2\right)\left(x-1\right)=0$

$x+2=0$

$x=-2$

Undefined domain x>0

$x-1=0$

$x=1$

Defined domain

To solve this problem, we'll carefully apply logarithmic properties:

• Step 1: Simplify the left-hand side:
  The left-hand side is given as $\log_64 \times \log_9x$. We simplify $\log_64$:
  $\log_64 = \frac{\log 4}{\log 6} = \frac{\log(2^2)}{\log 6} = \frac{2\log 2}{\log 6}$.
  Therefore, the left-hand side becomes $\frac{2\log 2}{\log 6} \times \log_9x$.
• Step 2: Simplify the right-hand side:
  The right-hand side is $(\log_6x^2 - \log_6x)(\log_92.5 + \log_91.6)$.
  First, simplify $\log_6x^2 - \log_6x = 2\log_6x - \log_6x = \log_6x$.
  For the other part, apply the product property: $\log_92.5 + \log_91.6 = \log_9(2.5 \times 1.6)$.
  Calculate $2.5 \times 1.6 = 4.0$, hence $\log_94$.
• Step 3: Equate and simplify:
  Now equate the simplified expressions: $\frac{2\log 2}{\log 6} \times \log_9x = \log_6x \cdot \log_94$.
  Change all logs to a common base (let's use natural log $\ln$) and solve:
• Step 4: Apply base conversion:
  $\log_9x = \frac{\ln x}{\ln 9}$, $\log_6x = \frac{\ln x}{\ln 6}$, and $\log_94 = \frac{\ln 4}{\ln 9}$.
• Step 5: Combine and solve:
  Perform algebraic manipulation and simplification:
  The equation becomes $\frac{2\ln 2}{\ln 6 \ln 9} \cdot \ln x = \frac{\ln x \cdot \ln 4}{\ln 6 \ln 9}$.
  Cancel $\ln x$ (non-zero due to $x > 0$) and solve for positive $x$.
• Conclude with the solution constraints:
  Given the properties and the domain involved, solution holds for all $0 < x$.

Therefore, the correct solution is: For all $0 < x$.

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

• Simplify the logarithmic expression using logarithmic identities.
• Substitute the simplified result back into the main expression and calculate its value.

Now, let's work through each step:

Step 1: Simplify the logarithmic expression. We'll simplify the parts involving $\log_7$ first, then those involving $\log_2$.

For the terms with $\log_7$:
- Convert $\log_7 x^n$ terms using the power rule: $\log_7 x^7 = 7 \log_7 x$, $\log_7 x^4 = 4 \log_7 x$, and $\log_7 x^3 = 3 \log_7 x$.
- The expression becomes $7 \log_7 x - 4 \log_7 x - 3 \log_7 x$.
- Simple arithmetic yields $0 \log_7 x$, which simplifies to $0$.

For the terms with $\log_2$:
- Similarly, $\log_2 y^n$ terms use the power rule: $\log_2 y^4 = 4 \log_2 y$, $\log_2 y^3 = 3 \log_2 y$, and $\log_2 y = 1 \log_2 y$.
- The expression is $4 \log_2 y - 3 \log_2 y - 1 \log_2 y$.
- Simple arithmetic gives $0 \log_2 y$, which also simplifies to $0$.

Step 2: Substitute these back into the original expression:

Original expression:
$\ln 4 \times (0 + 0) = \ln 4 \times 0 = 0$.

Therefore, the value of the expression is $\textbf{0}$.

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 $3\log_5 \frac{7}{24}$.