The Sum of Logarithms - Examples, Exercises and Solutions

To solve this problem, we will use the property of logarithms that allows us to combine the sum of two logarithms:

• Step 1: Identify the formula. We use the property $\log_b(x) + \log_b(y) = \log_b(x \cdot y)$ where both logarithms must have the same base.
• Step 2: Recognize the base. Here, both logarithms are in base 10: $\log_{10}3$ and $\log_{10}4$.
• Step 3: Apply the property. Add the two logarithms using the formula: $\log_{10}3 + \log_{10}4 = \log_{10}(3 \cdot 4)$.
• Step 4: Perform the multiplication. Compute $3 \cdot 4$ to get 12.
• Step 5: Express the result as a single logarithm: $\log_{10}12$.

Therefore, the expression $\log_{10}3 + \log_{10}4$ simplifies to $\log_{10}12$.

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

• Step 1: Identify the given expression as $\log_2 4 + \log_2 5$.
• Step 2: Use the sum of logarithms rule to simplify the expression.
• Step 3: Calculate the product and express the result.

Let's work through each step:

Step 1: We have $\log_2 4 + \log_2 5$ as our expression.

Step 2: Apply the sum of logarithms formula:

$\log_2 4 + \log_2 5 = \log_2 (4 \cdot 5)$

Step 3: Calculate the product:

$4 \times 5 = 20$

Thus, $\log_2 (4 \cdot 5) = \log_2 20$.

Therefore, the solution to the problem is $\log_2 20$.

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

• Step 1: Identify given logarithms and their base.
• Step 2: Employ the sum of logarithms property to combine the terms.
• Step 3: Calculate the resulting argument of the logarithm.

Now, let's work through each step:

Step 1: We have two logarithms: $\log_9 74$ and $\log_9 \frac{1}{2}$, sharing the base of $9$.

Step 2: Since the bases are the same, we use the sum property of logarithms:

$\log_9 74 + \log_9 \frac{1}{2} = \log_9 (74 \times \frac{1}{2})$.

Step 3: Calculate the product $74 \times \frac{1}{2}$:

$74 \times \frac{1}{2} = 37$.

So, we have:

$\log_9 (74 \times \frac{1}{2}) = \log_9 37$.

Therefore, the solution to the problem is $\log_9 37$.

$2\log_82=\log_82^2=\log_84$

$2\log_82+\log_83=\log_84+\log_83=$

$\log_84\cdot3=\log_812$

Where:

$3\log_49=\log_49^3=\log_4729$

y

$8\log_4\frac{1}{3}=\log_4\left(\frac{1}{3}\right)^8=$

$\log_4\frac{1}{3^8}=\log_4\frac{1}{6561}$

Therefore

$3\log_49+8\log_4\frac{1}{3}=$

$\log_4729+\log_4\frac{1}{6561}$

$\log_ax+\log_ay=\log_axy$

$\left(729\cdot\frac{1}{6561}\right)=\log_4\frac{1}{9}$

$\log_49^{-1}=-\log_49$

We break it down into parts

$\log_24=x$

$2^x=4$

$x=2$

$\log_39=x$

$3^x=9$

$x=2$

We substitute into the equation

$\frac{1}{2}\cdot2\log_38+2\log_37=$

$1\cdot\log_38+2\log_37=$

$\log_38+\log_37^2=$

$\log_38+\log_349=$

$\log_3\left(8\cdot49\right)=\log_3392$$x=2$

To solve this equation, we follow these steps:

• Step 1: Use the property of logarithms $\log_b a + \log_b c = \log_b (a \cdot c)$ to combine terms on the left-hand side of the equation.
• Step 2: Simplify the expression under the logarithm and solve for $x$.

Let's proceed through these steps:

Step 1: Rewrite the equation using logarithmic properties:
$\log_2 x + \log_2 \frac{x}{2} = \log_2 x + \log_2 x - \log_2 2$

This simplifies to:
$2 \log_2 x - 1 = 5$

Step 2: Solve the equation:
Add 1 to both sides:

Divide both sides by 2:

Now, convert the logarithmic equation to its exponential form:

Calculate $x$:

Therefore, the solution to the problem is $x = 8$.

To solve the given inequality $\log_4(7) + \log_4(2) \le \log_4(x)$, we will utilize the properties of logarithms:

• First, apply the logarithm sum property: $\log_4(7) + \log_4(2) = \log_4(7 \times 2) = \log_4(14)$.
• Now, the inequality becomes $\log_4(14) \le \log_4(x)$.
• Since the logarithm function is monotonically increasing when the base is greater than 1, we can simplify the inequality to $14 \le x$.

Therefore, the solution to the inequality is $x \ge 14$.

Therefore, the correct choice is $14 \le x$, which matches the given correct answer.

To solve the given logarithmic equation, let's proceed step-by-step:

• Step 1: Use the product rule of logarithms:
  Given the equation $\log_4 x + \log_4 (x+2) = 2$, apply the product rule to combine the logs:
  $\log_4 (x(x+2)) = 2$.
• Step 2: Convert the equation from logarithmic to exponential form:
  The equation becomes $x(x+2) = 4^2$, which simplifies to $x(x+2) = 16$.
• Step 3: Expand and rearrange the quadratic equation:
  We have $x^2 + 2x - 16 = 0$.
• Step 4: Solve the quadratic equation using the quadratic formula:
  The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = 1$, $b = 2$, and $c = -16$.
  Calculate the discriminant: $b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-16) = 4 + 64 = 68$.
  The solutions are given by:
  $x = \frac{-2 \pm \sqrt{68}}{2}$ which simplifies to $x = \frac{-2 \pm 2\sqrt{17}}{2}$.
  Thus, $x = -1 \pm \sqrt{17}$.
• Step 5: Check the solutions within the original equation's domain:
  Since $x$ must be greater than zero, $x = -1 - \sqrt{17}$ is invalid as it results in a negative value.
  Thus, the valid solution is $x = -1 + \sqrt{17}$.

Therefore, the solution to the problem is $x = -1 + \sqrt{17}$.

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

• Step 1: Combine the logarithms using the sum rule for logarithms.
• Step 2: Solve the resulting equation for $a$.
• Step 3: Ensure the solution meets domain restrictions.

Let's work through each step:

Step 1: We have the equation $\ln(a + 5) + \ln(a + 7) = 0$. Using the property of logarithms, combine the expressions:
$\ln(a+5) + \ln(a+7) = \ln((a+5)(a+7)) = 0$.

Step 2: Knowing $\ln((a+5)(a+7)) = 0$, use the exponential property that if $\ln(x) = 0$, then $x = 1$. Thus, set the expression inside the logarithm to 1:
$(a+5)(a+7) = 1$.

Now, expand and solve the equation:
$a^2 + 12a + 35 = 1$.
Rearrange this into a quadratic form:
$a^2 + 12a + 34 = 0$.

Step 3: Solve this quadratic equation using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1, b = 12, \text{ and } c = 34$:
$a = \frac{-12 \pm \sqrt{12^2 - 4 \times 1 \times 34}}{2 \times 1}$.

Calculate the discriminant:
$b^2 - 4ac = 144 - 136 = 8$.

Insert values back into the quadratic formula:
$a = \frac{-12 \pm \sqrt{8}}{2}$.
Simplify:
$a = \frac{-12 \pm 2\sqrt{2}}{2}$ = $-6 \pm \sqrt{2}$.

Given the domain restrictions: $a+5 > 0$ and $a+7 > 0$, we calculate the solutions:
The acceptable value is $-6 + \sqrt{2}$, since the domain restriction would invalidate another potential candidate.

Therefore, the solution to the problem is $-6 + \sqrt{2}$.

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

• Step 1: Combine the logarithms using the product rule.
• Step 2: Convert the logarithmic equation to an exponential equation.
• Step 3: Simplify and solve the quadratic equation.
• Step 4: Consider only solutions greater than 1.

Now, let's work through each step:
Step 1: Combine the logarithms using the product rule:
$\log(3x) + \log(x-1) = \log((3x)(x-1))$.
Step 2: Convert the logarithmic equation to an exponential equation:
$\log((3x)(x-1)) = 3 \Rightarrow (3x)(x-1) = 10^3$.
Step 3: Simplify the quadratic equation:
$(3x)(x-1) = 1000$:
$3x^2 - 3x = 1000$.
$\Rightarrow 3x^2 - 3x - 1000 = 0$.
Divide by 3 to simplify:
$x^2 - x - \frac{1000}{3} = 0$.
Solve this equation using the quadratic formula:
The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a = 1$, $b = -1$, and $c = -\frac{1000}{3}$.
Calculate the discriminant:\ $D = (-1)^2 - 4 \times 1 \times \left(-\frac{1000}{3}\right)\ = 1 + \frac{4000}{3} = \frac{4003}{3}$.
Now, calculate $x$:
$x = \frac{-(-1) \pm \sqrt{\frac{4003}{3}}}{2 \times 1}$.
$\Rightarrow x = \frac{1 \pm \sqrt{\frac{4003}{3}}}{2}$.
Calculating this gives approximately $x \approx 18.8$.
Step 4: Verify that $x > 1$ to be in the domain.
Since this is true, the valid solution is within the domain, confirming:
Therefore, the solution to the problem is $x = 18.8$.

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 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 $6$.

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 $\log_8 10$.