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Sagot :
Certainly! Let's solve the given equation step-by-step and verify which of the given logarithmic expressions are equivalent solutions.
We start with the given equation:
[tex]\[ 8 e^x - 5 = 0 \][/tex]
1. Isolate the exponential term:
[tex]\[ 8 e^x = 5 \][/tex]
2. Solve for [tex]\( e^x \)[/tex]:
[tex]\[ e^x = \frac{5}{8} \][/tex]
3. Take the natural logarithm of both sides:
[tex]\[ \ln(e^x) = \ln\left(\frac{5}{8}\right) \][/tex]
4. Simplify using properties of logarithms:
[tex]\[ x \ln(e) = \ln\left(\frac{5}{8}\right) \][/tex]
Since [tex]\(\ln(e) = 1\)[/tex], we have:
[tex]\[ x = \ln\left(\frac{5}{8}\right) \][/tex]
So we have:
[tex]\[ x = \ln\left(\frac{5}{8}\right) \][/tex]
Next, let's check the given options to see which ones are equivalent to [tex]\( x = \ln\left(\frac{5}{8}\right) \)[/tex]:
1. [tex]\( x = \ln(5) - \ln(8) \)[/tex]:
Using the property [tex]\(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)[/tex], we have:
[tex]\[\ln\left(\frac{5}{8}\right) = \ln(5) - \ln(8)\][/tex]
This is an equivalent expression.
2. [tex]\( x = \ln\left(\frac{5}{8}\right) \)[/tex]:
This is already our derived solution.
This is an equivalent expression.
3. [tex]\( x = \ln(5) + \ln(8) \)[/tex]:
Using the property [tex]\(\ln(ab) = \ln(a) + \ln(b)\)[/tex], we see:
[tex]\[\ln(5) + \ln(8) = \ln(40)\][/tex]
This is not an equivalent expression.
4. [tex]\( x = \frac{\ln(5)}{\ln(8)} \)[/tex]:
This form doesn't represent a valid manipulation of the given equation and has no direct logarithmic property to simplify into the correct form.
This is not an equivalent expression.
5. [tex]\( x = \ln\left(\frac{8}{5}\right) \)[/tex]:
Using the property [tex]\(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)[/tex], we have:
[tex]\[\ln\left(\frac{8}{5}\right) = \ln(8) - \ln(5)\][/tex]
This is not an equivalent expression because it represents the inverse ratio.
6. [tex]\( x = \ln(8) - \ln(5) \)[/tex]:
Using the property [tex]\(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)[/tex], we see:
[tex]\[\ln(8) - \ln(5) = \ln\left(\frac{8}{5}\right)\][/tex]
This is not an equivalent expression.
So, the correct choices are:
- [tex]\( x = \ln(5) - \ln(8) \)[/tex]
- [tex]\( x = \ln\left(\frac{5}{8}\right) \)[/tex]
The correct options are:
[tex]\[ \boxed{1 \text{ and } 2} \][/tex]
We start with the given equation:
[tex]\[ 8 e^x - 5 = 0 \][/tex]
1. Isolate the exponential term:
[tex]\[ 8 e^x = 5 \][/tex]
2. Solve for [tex]\( e^x \)[/tex]:
[tex]\[ e^x = \frac{5}{8} \][/tex]
3. Take the natural logarithm of both sides:
[tex]\[ \ln(e^x) = \ln\left(\frac{5}{8}\right) \][/tex]
4. Simplify using properties of logarithms:
[tex]\[ x \ln(e) = \ln\left(\frac{5}{8}\right) \][/tex]
Since [tex]\(\ln(e) = 1\)[/tex], we have:
[tex]\[ x = \ln\left(\frac{5}{8}\right) \][/tex]
So we have:
[tex]\[ x = \ln\left(\frac{5}{8}\right) \][/tex]
Next, let's check the given options to see which ones are equivalent to [tex]\( x = \ln\left(\frac{5}{8}\right) \)[/tex]:
1. [tex]\( x = \ln(5) - \ln(8) \)[/tex]:
Using the property [tex]\(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)[/tex], we have:
[tex]\[\ln\left(\frac{5}{8}\right) = \ln(5) - \ln(8)\][/tex]
This is an equivalent expression.
2. [tex]\( x = \ln\left(\frac{5}{8}\right) \)[/tex]:
This is already our derived solution.
This is an equivalent expression.
3. [tex]\( x = \ln(5) + \ln(8) \)[/tex]:
Using the property [tex]\(\ln(ab) = \ln(a) + \ln(b)\)[/tex], we see:
[tex]\[\ln(5) + \ln(8) = \ln(40)\][/tex]
This is not an equivalent expression.
4. [tex]\( x = \frac{\ln(5)}{\ln(8)} \)[/tex]:
This form doesn't represent a valid manipulation of the given equation and has no direct logarithmic property to simplify into the correct form.
This is not an equivalent expression.
5. [tex]\( x = \ln\left(\frac{8}{5}\right) \)[/tex]:
Using the property [tex]\(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)[/tex], we have:
[tex]\[\ln\left(\frac{8}{5}\right) = \ln(8) - \ln(5)\][/tex]
This is not an equivalent expression because it represents the inverse ratio.
6. [tex]\( x = \ln(8) - \ln(5) \)[/tex]:
Using the property [tex]\(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)[/tex], we see:
[tex]\[\ln(8) - \ln(5) = \ln\left(\frac{8}{5}\right)\][/tex]
This is not an equivalent expression.
So, the correct choices are:
- [tex]\( x = \ln(5) - \ln(8) \)[/tex]
- [tex]\( x = \ln\left(\frac{5}{8}\right) \)[/tex]
The correct options are:
[tex]\[ \boxed{1 \text{ and } 2} \][/tex]
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