Answer :
Sure, let's go through each part step by step using logarithmic properties.
### (i) Simplify Using Logarithmic Properties
Given:
[tex]\[ \log_6 \left(\frac{216^x}{1296^x}\right) \log_x 6 \][/tex]
We start by simplifying the expression inside the logarithm:
- Notice 216 and 1296 can be written in terms of 6:
- [tex]\( 216 = 6^3 \)[/tex]
- [tex]\( 1296 = 6^4 \)[/tex]
Rewrite the fraction:
[tex]\[ \frac{216^x}{1296^x} = \frac{(6^3)^x}{(6^4)^x} = \frac{6^{3x}}{6^{4x}} = 6^{3x-4x} = 6^{-x} \][/tex]
Now the original expression becomes:
[tex]\[ \log_6 (6^{-x}) \log_x 6 \][/tex]
Using the property of logarithms [tex]\(\log_b (b^a) = a\)[/tex]:
[tex]\[ \log_6 (6^{-x}) = -x \][/tex]
So we have:
[tex]\[ -x \log_x 6 \][/tex]
Thus, the simplified form is:
[tex]\[ -x \log_x 6 \][/tex]
### (ii) Condense the Complex Logarithm into a Single Term
Given:
[tex]\[ \log (x+1)^2 + \log (2x-1)^3 - \log (x)^2 - \log (2x-1)^4 + 6\log (x+1) \][/tex]
Using the properties of logarithms:
1. Combine logarithms with the same base:
[tex]\[ \log (x+1)^2 + 6 \log (x+1) = \log (x+1)^{2+6} = \log (x+1)^8 \][/tex]
[tex]\[ \log (2x-1)^3 - \log (2x-1)^4 = \log \left(\frac{(2x-1)^3}{(2x-1)^4}\right) = \log (2x-1)^{-1} \][/tex]
[tex]\[ -\log (x)^2 = \log (x)^{-2} \][/tex]
Combine these:
[tex]\[ \log (x+1)^8 + \log (2x-1)^{-1} + \log (x)^{-2} \][/tex]
Using the product property [tex]\( \log A + \log B = \log (A \cdot B) \)[/tex]:
[tex]\[ \log \left((x+1)^8 \cdot (2x-1)^{-1} \cdot (x)^{-2}\right) \][/tex]
Thus, the condensed form is:
[tex]\[ \log \left( \frac{(x+1)^8}{(2x-1) \cdot x^2} \right) \][/tex]
### (iii) Solve: [tex]\( 10 e^{2x-3} = 15 e^{5x-7} \)[/tex]
First, divide both sides by 10:
[tex]\[ e^{2x-3} = 1.5 e^{5x-7} \][/tex]
Take the natural logarithm (ln) of both sides:
[tex]\[ \ln(e^{2x-3}) = \ln(1.5e^{5x-7}) \][/tex]
Using the property [tex]\(\ln(e^a) = a\)[/tex] and [tex]\(\ln(ab) = \ln(a) + \ln(b)\)[/tex]:
[tex]\[ 2x - 3 = \ln(1.5) + \ln(e^{5x-7}) \][/tex]
Further simplifying gives:
[tex]\[ 2x - 3 = \ln(1.5) + 5x - 7 \][/tex]
Move terms involving [tex]\( x \)[/tex] to one side and constants to the other:
[tex]\[ 2x - 5x = \ln(1.5) - 7 + 3 \][/tex]
[tex]\[ -3x = \ln(1.5) - 4 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{4 - \ln(1.5)}{3} \][/tex]
Thus, the numerical value for [tex]\( x \)[/tex] is approximately:
[tex]\[ x \approx 1.1981782972972785 \][/tex]
So, the results are:
[tex]\[ \text{(i)} \; -x \log_x 6 \\ \text{(ii)} \; \log \left( \frac{(x+1)^8}{(2x-1) \cdot x^2} \right) \\ \text{(iii)} \; x \approx 1.1981782972972785 \][/tex]
### (i) Simplify Using Logarithmic Properties
Given:
[tex]\[ \log_6 \left(\frac{216^x}{1296^x}\right) \log_x 6 \][/tex]
We start by simplifying the expression inside the logarithm:
- Notice 216 and 1296 can be written in terms of 6:
- [tex]\( 216 = 6^3 \)[/tex]
- [tex]\( 1296 = 6^4 \)[/tex]
Rewrite the fraction:
[tex]\[ \frac{216^x}{1296^x} = \frac{(6^3)^x}{(6^4)^x} = \frac{6^{3x}}{6^{4x}} = 6^{3x-4x} = 6^{-x} \][/tex]
Now the original expression becomes:
[tex]\[ \log_6 (6^{-x}) \log_x 6 \][/tex]
Using the property of logarithms [tex]\(\log_b (b^a) = a\)[/tex]:
[tex]\[ \log_6 (6^{-x}) = -x \][/tex]
So we have:
[tex]\[ -x \log_x 6 \][/tex]
Thus, the simplified form is:
[tex]\[ -x \log_x 6 \][/tex]
### (ii) Condense the Complex Logarithm into a Single Term
Given:
[tex]\[ \log (x+1)^2 + \log (2x-1)^3 - \log (x)^2 - \log (2x-1)^4 + 6\log (x+1) \][/tex]
Using the properties of logarithms:
1. Combine logarithms with the same base:
[tex]\[ \log (x+1)^2 + 6 \log (x+1) = \log (x+1)^{2+6} = \log (x+1)^8 \][/tex]
[tex]\[ \log (2x-1)^3 - \log (2x-1)^4 = \log \left(\frac{(2x-1)^3}{(2x-1)^4}\right) = \log (2x-1)^{-1} \][/tex]
[tex]\[ -\log (x)^2 = \log (x)^{-2} \][/tex]
Combine these:
[tex]\[ \log (x+1)^8 + \log (2x-1)^{-1} + \log (x)^{-2} \][/tex]
Using the product property [tex]\( \log A + \log B = \log (A \cdot B) \)[/tex]:
[tex]\[ \log \left((x+1)^8 \cdot (2x-1)^{-1} \cdot (x)^{-2}\right) \][/tex]
Thus, the condensed form is:
[tex]\[ \log \left( \frac{(x+1)^8}{(2x-1) \cdot x^2} \right) \][/tex]
### (iii) Solve: [tex]\( 10 e^{2x-3} = 15 e^{5x-7} \)[/tex]
First, divide both sides by 10:
[tex]\[ e^{2x-3} = 1.5 e^{5x-7} \][/tex]
Take the natural logarithm (ln) of both sides:
[tex]\[ \ln(e^{2x-3}) = \ln(1.5e^{5x-7}) \][/tex]
Using the property [tex]\(\ln(e^a) = a\)[/tex] and [tex]\(\ln(ab) = \ln(a) + \ln(b)\)[/tex]:
[tex]\[ 2x - 3 = \ln(1.5) + \ln(e^{5x-7}) \][/tex]
Further simplifying gives:
[tex]\[ 2x - 3 = \ln(1.5) + 5x - 7 \][/tex]
Move terms involving [tex]\( x \)[/tex] to one side and constants to the other:
[tex]\[ 2x - 5x = \ln(1.5) - 7 + 3 \][/tex]
[tex]\[ -3x = \ln(1.5) - 4 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{4 - \ln(1.5)}{3} \][/tex]
Thus, the numerical value for [tex]\( x \)[/tex] is approximately:
[tex]\[ x \approx 1.1981782972972785 \][/tex]
So, the results are:
[tex]\[ \text{(i)} \; -x \log_x 6 \\ \text{(ii)} \; \log \left( \frac{(x+1)^8}{(2x-1) \cdot x^2} \right) \\ \text{(iii)} \; x \approx 1.1981782972972785 \][/tex]
