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Sagot :
To solve for [tex]\( x \)[/tex] given the equation:
[tex]\[ 2 e^{5x - 2} - 5 = 3 \][/tex]
we can follow these steps:
### Step 1: Isolate the exponential term
First, add 5 to both sides of the equation:
[tex]\[ 2 e^{5x - 2} - 5 + 5 = 3 + 5 \][/tex]
[tex]\[ 2 e^{5x - 2} = 8 \][/tex]
### Step 2: Solve for the exponential term
Next, divide both sides by 2:
[tex]\[ \frac{2 e^{5x - 2}}{2} = \frac{8}{2} \][/tex]
[tex]\[ e^{5x - 2} = 4 \][/tex]
### Step 3: Apply the natural logarithm
Take the natural logarithm ([tex]\(\ln\)[/tex]) of both sides to solve for [tex]\(x\)[/tex]:
[tex]\[ \ln(e^{5x - 2}) = \ln(4) \][/tex]
Recall that [tex]\(\ln(e^y) = y\)[/tex], so:
[tex]\[ 5x - 2 = \ln(4) \][/tex]
### Step 4: Solve for [tex]\(x\)[/tex]
Now, isolate [tex]\(x\)[/tex]:
[tex]\[ 5x = \ln(4) + 2 \][/tex]
[tex]\[ x = \frac{\ln(4) + 2}{5} \][/tex]
Let's compare this result to the given options.
### Given Options:
1. [tex]\(\frac{\ln 4 + 2}{5}\)[/tex]
2. [tex]\(\frac{\ln(-1) + 2}{5}\)[/tex] (Note: [tex]\(\ln(-1)\)[/tex] is undefined)
3. [tex]\(\frac{\ln 3 - \ln 2 + \ln 5 + 2}{5}\)[/tex]
4. [tex]\(1.2\)[/tex]
### Verify Each Option:
1. Option 1: [tex]\(\frac{\ln 4 + 2}{5}\)[/tex]
- This is the exact form we derived.
2. Option 2: [tex]\(\frac{\ln(-1) + 2}{5}\)[/tex]
- This is undefined because the natural logarithm of a negative number is not defined in the set of real numbers.
3. Option 3: [tex]\(\frac{\ln 3 - \ln 2 + \ln 5 + 2}{5}\)[/tex]
- This simplifies as follows:
[tex]\[ \frac{\ln 3 - \ln 2 + \ln 5 + 2}{5} = \frac{\ln \left(\frac{3 \cdot 5}{2}\right) + 2}{5} = \frac{\ln \left(\frac{15}{2}\right) + 2}{5} \][/tex]
This does not match our derived form.
4. Option 4: [tex]\(1.2\)[/tex]
- This is a numerical value. Upon closer inspection or calculation, it differs from our derived form.
### Conclusion
The correct solution aligns with the result derived from our steps and matches option 1:
[tex]\[ \boxed{\frac{\ln 4 + 2}{5}} \][/tex]
The numerical result of this calculation is approximately [tex]\(0.677\)[/tex].
[tex]\[ 2 e^{5x - 2} - 5 = 3 \][/tex]
we can follow these steps:
### Step 1: Isolate the exponential term
First, add 5 to both sides of the equation:
[tex]\[ 2 e^{5x - 2} - 5 + 5 = 3 + 5 \][/tex]
[tex]\[ 2 e^{5x - 2} = 8 \][/tex]
### Step 2: Solve for the exponential term
Next, divide both sides by 2:
[tex]\[ \frac{2 e^{5x - 2}}{2} = \frac{8}{2} \][/tex]
[tex]\[ e^{5x - 2} = 4 \][/tex]
### Step 3: Apply the natural logarithm
Take the natural logarithm ([tex]\(\ln\)[/tex]) of both sides to solve for [tex]\(x\)[/tex]:
[tex]\[ \ln(e^{5x - 2}) = \ln(4) \][/tex]
Recall that [tex]\(\ln(e^y) = y\)[/tex], so:
[tex]\[ 5x - 2 = \ln(4) \][/tex]
### Step 4: Solve for [tex]\(x\)[/tex]
Now, isolate [tex]\(x\)[/tex]:
[tex]\[ 5x = \ln(4) + 2 \][/tex]
[tex]\[ x = \frac{\ln(4) + 2}{5} \][/tex]
Let's compare this result to the given options.
### Given Options:
1. [tex]\(\frac{\ln 4 + 2}{5}\)[/tex]
2. [tex]\(\frac{\ln(-1) + 2}{5}\)[/tex] (Note: [tex]\(\ln(-1)\)[/tex] is undefined)
3. [tex]\(\frac{\ln 3 - \ln 2 + \ln 5 + 2}{5}\)[/tex]
4. [tex]\(1.2\)[/tex]
### Verify Each Option:
1. Option 1: [tex]\(\frac{\ln 4 + 2}{5}\)[/tex]
- This is the exact form we derived.
2. Option 2: [tex]\(\frac{\ln(-1) + 2}{5}\)[/tex]
- This is undefined because the natural logarithm of a negative number is not defined in the set of real numbers.
3. Option 3: [tex]\(\frac{\ln 3 - \ln 2 + \ln 5 + 2}{5}\)[/tex]
- This simplifies as follows:
[tex]\[ \frac{\ln 3 - \ln 2 + \ln 5 + 2}{5} = \frac{\ln \left(\frac{3 \cdot 5}{2}\right) + 2}{5} = \frac{\ln \left(\frac{15}{2}\right) + 2}{5} \][/tex]
This does not match our derived form.
4. Option 4: [tex]\(1.2\)[/tex]
- This is a numerical value. Upon closer inspection or calculation, it differs from our derived form.
### Conclusion
The correct solution aligns with the result derived from our steps and matches option 1:
[tex]\[ \boxed{\frac{\ln 4 + 2}{5}} \][/tex]
The numerical result of this calculation is approximately [tex]\(0.677\)[/tex].
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