Q. 1 - Calculus

Question

Solving exponential and logarithmic equations: Use rules of exponents and logarithms to solve each of the following equations.

$$\begin{gathered} a. 3{\left(1.2\right)}^x=500 \\ b. \frac{\ln \left(x+1\right)}{\ln \left(x-2\right)}=0 \\ c. \ln \left(x^2+x-5\right)=0 \\ d. \frac{2^x+1}{3^x-5}=0 \end{gathered}$$

Step-by-Step Solution

Answer

Part $\left(a\right)$: The value of $x$ for $3{\left(1.2\right)}^x=500$ is, $28.06$.

Part $\left(b\right)$: The value of $x$ for $\frac{\ln \left(x+1\right)}{\ln \left(x-2\right)}=0$ is, $0.$

Part $\left(c\right)$: The value of $x$ for $\ln \left(x^2+x-5\right)=0$ is, $2,-3$.

Part $\left(d\right)$: The value of $x$ for $\frac{2^x+1}{3^x-5}=0$ is undefined.

Part a Step 1 . Given information

$3{\left(1.2\right)}^x=500$.

Part a Step 2 . Now simplify the given expressions using the logarithmic rules.

$$\begin{gathered} 3{\left(1.2\right)}^x=500 \\ {\left(1.2\right)}^x=\frac{500}{3} \end{gathered}$$

Taking logarithm base of $1.2$ on both sides,

$$\begin{gathered} {\log }_{1.2}{\left(1.2\right)}^x={\log }_{1.2}\left(\frac{500}{3}\right) \\ x{\log }_{1.2}\left(1.2\right)={\log }_{1.2}\left(\frac{500}{3}\right) [\because {\log }_a{\left(b\right)}^m=m{\log }_a\left(b\right)] \\ x\left(1\right)=28.06 \\ x=28.06 \end{gathered}$$

Part b Step 1 . Given information

$\frac{\ln \left(x+1\right)}{\ln \left(x-2\right)}=0$.

Part b Step 2 . Use the exponential and logarithmic rules in the given expression.

$$\begin{gathered} \frac{\ln \left(x+1\right)}{\ln \left(x-2\right)}=0 \\ \ln \left(x+1\right)=0 \end{gathered}$$

As per the logarithmic properties,

 $$\begin{gathered} {\log }_a\left(y\right)=X \\ y=a^X \end{gathered}$$

So,

$$\begin{gathered} {\log }_e\left(x+1\right)=0 \\ x+1=e^0 \\ x+1=1 \\ x=1-1 \\ x=0 \end{gathered}$$

Part c Step 1 . Given information

$\ln \left(x^2+x-5\right)=0$.

Part c Step 2 . Use the logarithmic and exponential rules to simplify the given expressions.

$$\begin{gathered} \ln \left(x^2+x-5\right)=0 \\ {\log }_e\left(x^2+x-5\right)=0 \end{gathered}$$

As per the logarithmic properties,

$$\begin{gathered} {\log }_a\left(y\right)=X \\ y=a^X \end{gathered}$$

So,

$$\begin{gathered} x^2+x-5=e^0 \\ {x}^2+x-5=1 \\ {x}^2+x-5-1=0 \\ {x}^2+x-6=0 \\ \left(x-2\right)\left(x+3\right)=0 \\ x=2,-3 \end{gathered}$$

Part d Step 1 . Given information

$\frac{2^x+1}{3^x-5}=0$.

Part d Step 2 . Use the logarithmic and exponential rules to evaluate the given expression.

$$\begin{gathered} \frac{2^x+1}{3^x-5}=0 \\ {2}^x+1=0 \\ {2}^x=-1 \end{gathered}$$

Taking logarithm base of $2$ on both sides,

${\log }_2\left(2^x\right)={\log }_2\left(-1\right)$ , which is undefined.

Since the logarithm of a negative number is undefined.