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Sagot :
Let's solve the system of equations step-by-step:
1. Solve the first equation [tex]\(12^{x+5} = 8\)[/tex]:
Start by isolating the exponential term. Given
[tex]\[ 12^{x+5} = 8 \][/tex]
Take the natural logarithm (ln) on both sides to move the exponent:
[tex]\[ \ln(12^{x+5}) = \ln(8) \][/tex]
Use the logarithm power rule [tex]\(\ln(a^b) = b \ln(a)\)[/tex]:
[tex]\[ (x + 5) \ln(12) = \ln(8) \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ x + 5 = \frac{\ln(8)}{\ln(12)} \][/tex]
Subtract 5 from both sides to isolate [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\ln(8)}{\ln(12)} - 5 \][/tex]
Note that [tex]\(\ln(8)\)[/tex] could be expressed in terms of [tex]\(\ln(2)\)[/tex] since [tex]\(8 = 2^3\)[/tex]:
[tex]\[ \ln(8) = \ln(2^3) = 3 \ln(2) \][/tex]
However, we'll keep it in the form involving [tex]\(\ln(8)\)[/tex] explicitly:
[tex]\[ x = \frac{\ln(8)}{\ln(12)} - 5 \][/tex]
Rewriting the solution, noticing that [tex]\(\frac{\ln(8)}{\ln(12)}\)[/tex] simplifies to [tex]\(\log_{12}(8)\)[/tex], we get:
[tex]\[ x = \log_{12}(8) - 5 \][/tex]
Alternatively:
[tex]\[ x = \frac{\log(8)}{\log(12)} - 5 \][/tex]
2. Solve the second equation [tex]\(e^y = 7\)[/tex]:
Take the natural logarithm on both sides to solve for [tex]\(y\)[/tex]:
[tex]\[ \ln(e^y) = \ln(7) \][/tex]
Use the rule [tex]\(\ln(e^b) = b\)[/tex]:
[tex]\[ y = \ln(7) \][/tex]
Summary:
- For [tex]\(12^{x+5} = 8\)[/tex],
[tex]\[ x = \frac{\ln(8)}{\ln(12)} - 5 = -\frac{\log(31104)}{\log(12)} \][/tex]
- For [tex]\(e^y = 7\)[/tex],
[tex]\[ y = \ln(7) \][/tex]
Thus, the solutions are:
[tex]\[ x = -\log_{12}(31104) \][/tex]
[tex]\[ y = \ln(7) \][/tex]
1. Solve the first equation [tex]\(12^{x+5} = 8\)[/tex]:
Start by isolating the exponential term. Given
[tex]\[ 12^{x+5} = 8 \][/tex]
Take the natural logarithm (ln) on both sides to move the exponent:
[tex]\[ \ln(12^{x+5}) = \ln(8) \][/tex]
Use the logarithm power rule [tex]\(\ln(a^b) = b \ln(a)\)[/tex]:
[tex]\[ (x + 5) \ln(12) = \ln(8) \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ x + 5 = \frac{\ln(8)}{\ln(12)} \][/tex]
Subtract 5 from both sides to isolate [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\ln(8)}{\ln(12)} - 5 \][/tex]
Note that [tex]\(\ln(8)\)[/tex] could be expressed in terms of [tex]\(\ln(2)\)[/tex] since [tex]\(8 = 2^3\)[/tex]:
[tex]\[ \ln(8) = \ln(2^3) = 3 \ln(2) \][/tex]
However, we'll keep it in the form involving [tex]\(\ln(8)\)[/tex] explicitly:
[tex]\[ x = \frac{\ln(8)}{\ln(12)} - 5 \][/tex]
Rewriting the solution, noticing that [tex]\(\frac{\ln(8)}{\ln(12)}\)[/tex] simplifies to [tex]\(\log_{12}(8)\)[/tex], we get:
[tex]\[ x = \log_{12}(8) - 5 \][/tex]
Alternatively:
[tex]\[ x = \frac{\log(8)}{\log(12)} - 5 \][/tex]
2. Solve the second equation [tex]\(e^y = 7\)[/tex]:
Take the natural logarithm on both sides to solve for [tex]\(y\)[/tex]:
[tex]\[ \ln(e^y) = \ln(7) \][/tex]
Use the rule [tex]\(\ln(e^b) = b\)[/tex]:
[tex]\[ y = \ln(7) \][/tex]
Summary:
- For [tex]\(12^{x+5} = 8\)[/tex],
[tex]\[ x = \frac{\ln(8)}{\ln(12)} - 5 = -\frac{\log(31104)}{\log(12)} \][/tex]
- For [tex]\(e^y = 7\)[/tex],
[tex]\[ y = \ln(7) \][/tex]
Thus, the solutions are:
[tex]\[ x = -\log_{12}(31104) \][/tex]
[tex]\[ y = \ln(7) \][/tex]
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