Answer :
To solve the exponential equation [tex]\( e^{2x+1} = 15 \)[/tex], we need to isolate [tex]\( x \)[/tex].
Here are the steps to find the solution in terms of logarithms, and to get the numerical value correct to four decimal places:
1. Isolate the exponent:
[tex]\[ e^{2x+1} = 15 \][/tex]
2. Take the natural logarithm of both sides:
[tex]\[ \ln(e^{2x+1}) = \ln(15) \][/tex]
3. Use the property of logarithms [tex]\(\ln(e^y) = y\)[/tex]:
[tex]\[ 2x + 1 = \ln(15) \][/tex]
4. Isolate [tex]\( x \)[/tex]:
Subtract 1 from both sides:
[tex]\[ 2x = \ln(15) - 1 \][/tex]
Then, divide both sides by 2:
[tex]\[ x = \frac{\ln(15) - 1}{2} \][/tex]
In logarithmic form, the solution is:
[tex]\[ x = \frac{\ln(15) - 1}{2} \][/tex]
To find the numerical value,
use [tex]\(\ln(15)\)[/tex],
Numerically evaluating the expression:
[tex]\[ x \approx 0.8540 \][/tex]
So, the solution correct to four decimal places is:
[tex]\[ x = 0.8540 \][/tex]
Here are the steps to find the solution in terms of logarithms, and to get the numerical value correct to four decimal places:
1. Isolate the exponent:
[tex]\[ e^{2x+1} = 15 \][/tex]
2. Take the natural logarithm of both sides:
[tex]\[ \ln(e^{2x+1}) = \ln(15) \][/tex]
3. Use the property of logarithms [tex]\(\ln(e^y) = y\)[/tex]:
[tex]\[ 2x + 1 = \ln(15) \][/tex]
4. Isolate [tex]\( x \)[/tex]:
Subtract 1 from both sides:
[tex]\[ 2x = \ln(15) - 1 \][/tex]
Then, divide both sides by 2:
[tex]\[ x = \frac{\ln(15) - 1}{2} \][/tex]
In logarithmic form, the solution is:
[tex]\[ x = \frac{\ln(15) - 1}{2} \][/tex]
To find the numerical value,
use [tex]\(\ln(15)\)[/tex],
Numerically evaluating the expression:
[tex]\[ x \approx 0.8540 \][/tex]
So, the solution correct to four decimal places is:
[tex]\[ x = 0.8540 \][/tex]