Find the eventual solution(s) of the following equations.
3^x=24
Solving a^x=b requires taking the log (with any base) of both sides. Using natural log, the equation becomes:
\ln\left(a^x\right)=\ln\left(b\right)
x\cdot \ln\left(a\right)=\ln\left(b\right)
Therefore, the solution to 3^x=24 is found by taking the natural log of both sides:
x\cdot \ln\left(3\right)=\ln\left(24\right)
Dividing both sides by \ln\left(3\right) gives:
x=\dfrac{\ln\left(24\right)}{\ln\left(3\right)}
The solution is x=\dfrac{\ln\left(24\right)}{\ln\left(3\right)}
2^x=19
Solving a^x=b requires taking the log (with any base) of both sides. Using natural log, the equation becomes:
\ln\left(a^x\right)=\ln\left(b\right)
x\cdot \ln\left(a\right)=\ln\left(b\right)
Therefore, the solution to 2^x=19 is found by taking the natural log of both sides:
x\cdot \ln\left(2\right)=\ln\left(19\right)
Dividing both sides by \ln\left(2\right) gives:
x=\dfrac{\ln\left(19\right)}{\ln\left(2\right)}
The solution is x=\dfrac{\ln\left(19\right)}{\ln\left(2\right)}.
10^x=73
Solving a^x=b requires taking the log (with any base) of both sides. Using natural log, the equation becomes:
\ln\left(a^x\right)=\ln\left(b\right)
x\cdot \ln\left(a\right)=\ln\left(b\right)
Therefore, the solution to 10^x=73 is found by taking the natural log of both sides:
x\cdot \ln\left(10\right)=\ln\left(73\right)
Dividing both sides by \ln\left(10\right) gives:
x=\dfrac{\ln\left(73\right)}{\ln\left(10\right)}
The solution is x=\dfrac{\ln\left(73\right)}{\ln\left(10\right)}.
\left(\dfrac{1}{2}\right)^x=70
Solving a^x=b requires taking the log (with any base) of both sides. Using natural log, the equation becomes:
\ln\left(a^x\right)=\ln\left(b\right)
x\cdot \ln\left(a\right)=\ln\left(b\right)
Therefore, the solution to \left(\dfrac{1}{2}\right)^x=70 is found by taking the natural log of both sides:
x\cdot \ln\left(\dfrac{1}{2}\right)=\ln\left(70\right)
Dividing both sides by \ln\left(\dfrac{1}{2}\right) gives:
x=\dfrac{\ln\left(70\right)}{\ln\left(\dfrac{1}{2}\right)}=-\dfrac{\ln\left(70\right)}{\ln\left(2\right)}
The solution is x=-\dfrac{\ln\left(70\right)}{\ln\left(2\right)}.
\left(\dfrac{1}{3}\right)^x=81
Solving a^x=b requires taking the log (with any base) of both sides. Using natural log, the equation becomes:
\ln\left(a^x\right)=\ln\left(b\right)
x\cdot \ln\left(a\right)=\ln\left(b\right)
Therefore, the solution to 3^{-x}=81 is found by taking the natural log of both sides:
-x\cdot \ln\left(3\right)=\ln\left(81\right)
Dividing both sides by -\ln\left(3\right) gives:
x=-\dfrac{\ln\left(81\right)}{\ln\left(3\right)}=-4
The solution is x=-4.
\pi^x=50
Solving a^x=b requires taking the log (with any base) of both sides. Using natural log, the equation becomes:
\ln\left(a^x\right)=\ln\left(b\right)
x\cdot \ln\left(a\right)=\ln\left(b\right)
Therefore, the solution to \pi^x=50 is found by taking the natural log of both sides:
x\cdot \ln\left(\pi\right)=\ln\left(50\right)
Dividing both sides by \ln\left(\pi\right) gives:
x=\dfrac{\ln\left(50\right)}{\ln\left(\pi\right)}
The solution is x=\dfrac{\ln\left(50\right)}{\ln\left(\pi\right)}.
10^x=99
Solving a^x=b requires taking the log (with any base) of both sides. Using natural log, the equation becomes:
\ln\left(a^x\right)=\ln\left(b\right)
x\cdot \ln\left(a\right)=\ln\left(b\right)
Therefore, the solution to 10^x=99 is found by taking the natural log of both sides:
x\cdot \ln\left(10\right)=\ln\left(99\right)
Dividing both sides by \ln\left(3\right) gives:
x=\dfrac{\ln\left(99\right)}{\ln\left(10\right)}
The solution is x=\dfrac{\ln\left(99\right)}{\ln\left(10\right)}.