Answer :
Let's solve the given math problems step-by-step.
#### c) Rewrite [tex]\(\log _4(6 x)+\frac{1}{2}-\log _4(y z)\)[/tex]
To simplify the expression [tex]\(\log_4(6x) + \frac{1}{2} - \log_4(yz)\)[/tex], use the properties of logarithms, particularly the logarithm subtraction rule and the logarithm addition rule:
1. Subtraction Rule for Logarithms:
[tex]\[ \log_b(A) - \log_b(B) = \log_b \left( \frac{A}{B} \right) \][/tex]
Apply this rule to get:
[tex]\[ \log_4(6x) - \log_4(yz) = \log_4 \left( \frac{6x}{yz} \right) \][/tex]
2. Combine with Addition:
Use the property of adding a constant: When you add [tex]\(\frac{1}{2}\)[/tex] to the logarithm, you can express it as multiplying the argument of the logarithm by the base raised to [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \log_4 \left( \frac{6x}{yz} \right) + \frac{1}{2} = \log_4 \left( \left(\frac{6x}{yz}\right) \cdot 4^{\frac{1}{2}} \right) \][/tex]
Since [tex]\(4^{\frac{1}{2}} = 2\)[/tex], this simplifies to:
[tex]\[ \log_4 \left( \frac{6x \cdot 2}{yz} \right) = \log_4 \left( \frac{12x}{yz} \right) \][/tex]
Thus, the simplified expression is:
[tex]\[ \log_4 \left( \frac{12x}{yz} \right) \][/tex]
#### d) Rewrite [tex]\(\log _5 a\)[/tex]
The given expression [tex]\(\log_5 a\)[/tex] is already in its simplest form. Nothing needs to be done here:
[tex]\[ \log_5 a \][/tex]
#### 10. Solve the following equations:
a) [tex]\(\log \sqrt{a} - \log_{\sqrt{5}} a = 2\)[/tex]
1. Change of Base Formula:
Use change of base formula for [tex]\(\log_{\sqrt{5}} a\)[/tex]:
[tex]\[ \log_{\sqrt{5}} a = \frac{\log a}{\log \sqrt{5}} \][/tex]
2. Simplify [tex]\(\log \sqrt{5}\)[/tex]:
[tex]\[ \log \sqrt{5} = \frac{1}{2} \log 5 \][/tex]
So:
[tex]\[ \log_{\sqrt{5}} a = \frac{\log a}{\frac{1}{2} \log 5} = \frac{2 \log a}{\log 5} \][/tex]
3. Substitute and simplify:
[tex]\[ \log \sqrt{a} - \frac{2 \log a}{\log 5} = 2 \][/tex]
Since [tex]\(\log \sqrt{a} = \frac{1}{2} \log a\)[/tex]:
[tex]\[ \frac{1}{2} \log a - \frac{2 \log a}{\log 5} = 2 \][/tex]
Let [tex]\(\log a = x\)[/tex]:
[tex]\[ \frac{1}{2} x - \frac{2x}{\log 5} = 2 \][/tex]
Multiply by 2 [tex]\(\log 5\)[/tex] to clear the fraction:
[tex]\[ (\log 5) x - 4 x = 4 \log 5 \][/tex]
Factor the [tex]\(x\)[/tex] term:
[tex]\[ x (\log 5 - 4) = 4 \log 5 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{4 \log 5}{\log 5 - 4} \][/tex]
Since [tex]\(x = \log a\)[/tex]:
[tex]\[ \log a = \frac{4 \log 5}{\log 5 - 4} \][/tex]
Convert back to exponential form:
[tex]\[ a = 10^{\frac{4 \log 5}{\log 5 - 4}} \][/tex]
b) [tex]\(2 \log _3\)[/tex]
The given expression is incomplete. It is missing either the variable or another part of the expression.
c) [tex]\(\ln(c + 3) + \ln(4 - c) = \ln(2 - 2c)\)[/tex]
1. Combine Logarithms:
Use the logarithm property that adds two logs:
[tex]\[ \ln(c + 3) + \ln(4 - c) = \ln((c + 3)(4 - c)) \][/tex]
So the equation becomes:
[tex]\[ \ln((c + 3)(4 - c)) = \ln(2 - 2c) \][/tex]
2. Since [tex]\(\ln A = \ln B\)[/tex]:
If the natural logs are equal, set their arguments equal:
[tex]\[ (c + 3)(4 - c) = 2 - 2c \][/tex]
3. Expand and simplify:
Expand the left side:
[tex]\[ 4c + 12 - c^2 - 3c = 2 - 2c \][/tex]
Combine like terms:
[tex]\[ c + 12 - c^2 = 2 - 2c \][/tex]
Rearrange into a standard quadratic equation form:
[tex]\[ -c^2 + 3c + 10 = 0 \][/tex]
Multiply by -1 to simplify:
[tex]\[ c^2 - 3c - 10 = 0 \][/tex]
4. Solve the quadratic equation:
Factor the quadratic:
[tex]\[ (c - 5)(c + 2) = 0 \][/tex]
So, the solutions are:
[tex]\[ c = 5 \text{ or } c = -2 \][/tex]
5. Verify the solutions in the original equation:
- For [tex]\(c = 5\)[/tex]:
[tex]\[ \ln(5 + 3) + \ln(4 - 5) = \ln(2 - 2 \cdot 5) \][/tex]
[tex]\[ \ln(8) + \ln(-1) = \ln(-8) \quad \text{(not valid, since \(\ln \) of a negative number is undefined)} \][/tex]
- For [tex]\(c = -2\)[/tex]:
[tex]\[ \ln(-2 + 3) + \ln(4 - (-2)) = \ln(2 - 2 \cdot (-2)) \][/tex]
[tex]\[ \ln(1) + \ln(6) = \ln(6) \][/tex]
[tex]\[ 0 + \ln(6) = \ln(6) \quad \text{(valid)} \][/tex]
Hence, the valid solution is:
[tex]\[ c = -2 \][/tex]
d) [tex]\(\log_2 d\)[/tex]
The expression [tex]\(\log_2 d\)[/tex] is already in its simplest form. Like previously mentioned, no further simplification or solving is possible without additional context or more information.
Feel free to ask if you need any further clarification or another topic!
#### c) Rewrite [tex]\(\log _4(6 x)+\frac{1}{2}-\log _4(y z)\)[/tex]
To simplify the expression [tex]\(\log_4(6x) + \frac{1}{2} - \log_4(yz)\)[/tex], use the properties of logarithms, particularly the logarithm subtraction rule and the logarithm addition rule:
1. Subtraction Rule for Logarithms:
[tex]\[ \log_b(A) - \log_b(B) = \log_b \left( \frac{A}{B} \right) \][/tex]
Apply this rule to get:
[tex]\[ \log_4(6x) - \log_4(yz) = \log_4 \left( \frac{6x}{yz} \right) \][/tex]
2. Combine with Addition:
Use the property of adding a constant: When you add [tex]\(\frac{1}{2}\)[/tex] to the logarithm, you can express it as multiplying the argument of the logarithm by the base raised to [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \log_4 \left( \frac{6x}{yz} \right) + \frac{1}{2} = \log_4 \left( \left(\frac{6x}{yz}\right) \cdot 4^{\frac{1}{2}} \right) \][/tex]
Since [tex]\(4^{\frac{1}{2}} = 2\)[/tex], this simplifies to:
[tex]\[ \log_4 \left( \frac{6x \cdot 2}{yz} \right) = \log_4 \left( \frac{12x}{yz} \right) \][/tex]
Thus, the simplified expression is:
[tex]\[ \log_4 \left( \frac{12x}{yz} \right) \][/tex]
#### d) Rewrite [tex]\(\log _5 a\)[/tex]
The given expression [tex]\(\log_5 a\)[/tex] is already in its simplest form. Nothing needs to be done here:
[tex]\[ \log_5 a \][/tex]
#### 10. Solve the following equations:
a) [tex]\(\log \sqrt{a} - \log_{\sqrt{5}} a = 2\)[/tex]
1. Change of Base Formula:
Use change of base formula for [tex]\(\log_{\sqrt{5}} a\)[/tex]:
[tex]\[ \log_{\sqrt{5}} a = \frac{\log a}{\log \sqrt{5}} \][/tex]
2. Simplify [tex]\(\log \sqrt{5}\)[/tex]:
[tex]\[ \log \sqrt{5} = \frac{1}{2} \log 5 \][/tex]
So:
[tex]\[ \log_{\sqrt{5}} a = \frac{\log a}{\frac{1}{2} \log 5} = \frac{2 \log a}{\log 5} \][/tex]
3. Substitute and simplify:
[tex]\[ \log \sqrt{a} - \frac{2 \log a}{\log 5} = 2 \][/tex]
Since [tex]\(\log \sqrt{a} = \frac{1}{2} \log a\)[/tex]:
[tex]\[ \frac{1}{2} \log a - \frac{2 \log a}{\log 5} = 2 \][/tex]
Let [tex]\(\log a = x\)[/tex]:
[tex]\[ \frac{1}{2} x - \frac{2x}{\log 5} = 2 \][/tex]
Multiply by 2 [tex]\(\log 5\)[/tex] to clear the fraction:
[tex]\[ (\log 5) x - 4 x = 4 \log 5 \][/tex]
Factor the [tex]\(x\)[/tex] term:
[tex]\[ x (\log 5 - 4) = 4 \log 5 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{4 \log 5}{\log 5 - 4} \][/tex]
Since [tex]\(x = \log a\)[/tex]:
[tex]\[ \log a = \frac{4 \log 5}{\log 5 - 4} \][/tex]
Convert back to exponential form:
[tex]\[ a = 10^{\frac{4 \log 5}{\log 5 - 4}} \][/tex]
b) [tex]\(2 \log _3\)[/tex]
The given expression is incomplete. It is missing either the variable or another part of the expression.
c) [tex]\(\ln(c + 3) + \ln(4 - c) = \ln(2 - 2c)\)[/tex]
1. Combine Logarithms:
Use the logarithm property that adds two logs:
[tex]\[ \ln(c + 3) + \ln(4 - c) = \ln((c + 3)(4 - c)) \][/tex]
So the equation becomes:
[tex]\[ \ln((c + 3)(4 - c)) = \ln(2 - 2c) \][/tex]
2. Since [tex]\(\ln A = \ln B\)[/tex]:
If the natural logs are equal, set their arguments equal:
[tex]\[ (c + 3)(4 - c) = 2 - 2c \][/tex]
3. Expand and simplify:
Expand the left side:
[tex]\[ 4c + 12 - c^2 - 3c = 2 - 2c \][/tex]
Combine like terms:
[tex]\[ c + 12 - c^2 = 2 - 2c \][/tex]
Rearrange into a standard quadratic equation form:
[tex]\[ -c^2 + 3c + 10 = 0 \][/tex]
Multiply by -1 to simplify:
[tex]\[ c^2 - 3c - 10 = 0 \][/tex]
4. Solve the quadratic equation:
Factor the quadratic:
[tex]\[ (c - 5)(c + 2) = 0 \][/tex]
So, the solutions are:
[tex]\[ c = 5 \text{ or } c = -2 \][/tex]
5. Verify the solutions in the original equation:
- For [tex]\(c = 5\)[/tex]:
[tex]\[ \ln(5 + 3) + \ln(4 - 5) = \ln(2 - 2 \cdot 5) \][/tex]
[tex]\[ \ln(8) + \ln(-1) = \ln(-8) \quad \text{(not valid, since \(\ln \) of a negative number is undefined)} \][/tex]
- For [tex]\(c = -2\)[/tex]:
[tex]\[ \ln(-2 + 3) + \ln(4 - (-2)) = \ln(2 - 2 \cdot (-2)) \][/tex]
[tex]\[ \ln(1) + \ln(6) = \ln(6) \][/tex]
[tex]\[ 0 + \ln(6) = \ln(6) \quad \text{(valid)} \][/tex]
Hence, the valid solution is:
[tex]\[ c = -2 \][/tex]
d) [tex]\(\log_2 d\)[/tex]
The expression [tex]\(\log_2 d\)[/tex] is already in its simplest form. Like previously mentioned, no further simplification or solving is possible without additional context or more information.
Feel free to ask if you need any further clarification or another topic!
