Precalculus

Expand the following expressions using one or more of the above rules:

$(a)$ $\log{x^y}$

$(b)$ $\log{\frac{5x}{y^3}}$

$(c)$ $\ln{\frac{(x+1)}{3z}}$

$(d)$ $\ln{\frac{\sqrt{x^2+3}}{x-1}}$

Solution:

$(a)$ $\log{x^y} = y\log{x}$

$(b)$ $\log{\frac{5x}{y^3}} = \log{5} + \log{x} - 3\log{y}$

$(c)$ $\ln{\frac{(x+1)}{3z}} = \ln{(x+1)}- \ln{3} - \ln{z}$

$(d)$ $\ln{\frac{\sqrt{x^2+3}}{x-1}} = \frac{1}{2} \ln{(x^2+3)} - \ln{(x-1)}$

Rewrite the following expressions as a single logarithmic expression with unit coefficient:

$(a)$ $3\log{x} + 2\log{y} - \log{z}$

$(b)$ $\log{(x-2)} - 4\log{(x^2-2x+4)}$

$(c)$ $\frac{1}{2}\ln{x} -\ln{(x+5)}$

$(d)$ $-\frac{1}{3}\ln{(x^2+x-4)} + 2\ln{y}$

Solution:

$(a)$ $3\log{x} + 2\log{y} - \log{z} = \log{\left(\frac{x^3y^2}{z}\right)}$

$(b)$ $\log{(x-2)} - 4\log{(x^2-2x+4)} = \log{\left(\frac{(x-2)}{(x^2-2x+4)^4}\right)}$

$(c)$ $\frac{1}{2}\ln{x} -\ln{(x+5)} = \ln{\left(\frac{\sqrt{x}}{(x+5)}\right)}$

$(d)$ $-\frac{1}{3}\ln{(x^2+x-4)} + 2\ln{y} = \ln{\left(\frac{y^2}{\sqrt[3]{x^2+x-4}}\right)}$

Solve the exponential equation $8^{(2x-1)} = 2^{3x}$

Solution:

Observing the bases on both sides, we note that the equation can be rewritten so that they have the same base.

\begin{eqnarray} 8^{(2x-1)} &=& 2^{3x}\\ \left(2^3\right)^{(2x-1)} &=& 2^{3x}\\ 2^{(3(2x-1))} &=& 2^{3x}\\ 2^{(6x-3)} &=& 2^{3x}\\ 6x-3 &=& 3x\\ 3x &=& 3\\ x &=& 1\\ \end{eqnarray}
Check: Substituting $x=1$ in the equation, we have \begin{eqnarray} 8^{(2-1)} &=& 2^{3}\\ 8^1 &=& 2^3\\ 8 &=& 8\\ \end{eqnarray}
Hence the solution is $x=1$.

Solve the exponential equation $32\left(\frac{1}{4}\right)^{\frac{x}{3}} = 2$

Solution:

This equation can be rewritten so that both sides have base $2$: \begin{eqnarray} 32\left(\frac{1}{4}\right)^{\frac{x}{3}} &=& 2\\ \left(2^5\right)\left(2^{-\frac{2x}{3}}\right)\\ 2^{(5-\frac{2x}{3})} &=& 2^1\\ 5-\frac{2x}{3} &=& 1\\ 4 &=& \frac{2x}{3}\\ 12 &=& 2x\\ 6 &=& x\\ \end{eqnarray}
Check: \begin{eqnarray} 32\left(\frac{1}{4}\right)^{\frac{x}{3}} &=& 2\\ 32\left(\frac{1}{4}\right)^{\frac{6}{3}} &=& 2\\ 32\left(\frac{1}{4}\right)^{2} &=& 2\\ \left(2^5\right)\left(2^{-4}\right) &=& 2\\ 2^{5-4} &=& 2\\ 2^1 &=& 2\\ \end{eqnarray}
Hence the solution is $x=6$

Solve the equation: $\frac{e^x - e^{-x}}{2} = 4$

Solution:

This equation can be solved by performing the following steps:

1. Step 1: Multiply both sides by $2$.
2. Step 2: Multiply both sides by $e^x$.
3. Step 3: Substitute $e^x = m$ and simplify. The resulting equations is a quadratic equation in $m$.
4. Step 4: Solve the quadratic equation in $m$ to get value(s) for $e^x$.
5. Step 5: Solve for x using rewriting the exponential equation as logarithmic equation.
6. Step 6: As a logarithmic function is defined only for positive real numbers, any negative or zero solution has to be discarded as extraneous. So, the final solution would be only the positive real number solutions got from step 5.

\begin{eqnarray} \frac{e^x - e^{-x}}{2} &=& 4\\ e^x - e^{-x} &=& 8\\ e^{2x} - 1 &=& 8e^x\\ e^{2x} - 8e^x - 1 &=& 0\\ m^2 - 8m - 1 &=& 0\\ m = 8.1231 \quad &or& \quad m = -0.1231\\ e^x = 8.1231 \quad &or& \quad e^x = -0.1231\\ \end{eqnarray}

Now, $e^x = 8.1231$ gives $x = \ln{8.1231}$, which gives $x = 2.0947$. But the second value $e^x = -0.1231$ has no solution as it is extraneous. Hence the solution to this equation is $x = 2.9=0947$

Solve the equation $\log_{4}{(x+1)} = 2$

Solution:

This equation is easily solved by rewriting it as exponential equation: \begin{eqnarray} \log_{4}{(x+1)} &=& 2\\ x + 1 &=& 4^2\\ x+1 &=& 16\\ x &=& 15\\ \end{eqnarray}
In this problem, substituting $x=15$ would make the expression $(x+1)$ positive, and so the equation is well-defined for this value. Hence we can conclude that the solution is $x=15$.

Solve the equation $\log{x} -\frac{1}{2}\log{(x+4)} = 1$

Solution:

Here, we need to rewrite the left side as a single logarithm with unit coefficient and then solve: \begin{eqnarray} \log{x} -\frac{1}{2}\log{(x+4)} &=& 1\\ \log{x} - \log{\sqrt{(x+4)}} &=& 1\\ \log{\left(\frac{x}{\sqrt{(x+4)}}\right)} &=& 1\\ \frac{x}{\sqrt{x+4}} &=& 10^1\\ x^2-100x-400 &=& 0\\ x = 103.85 \quad &=& \quad x = -3.85\\ \end{eqnarray}
Substituting each solution in the original equation proves that $x = -3.85$ is extraneous. Hence, solution for this equation is $x = 103.85$.

Solve: $\ln{(x-3)} + \ln{(x+4)} = 3\ln{2}$

Solution:

First, we need to rewrite each side as a single logarithm with unit coefficient. Then solve the equation obtained by setting the expressions inside the logarithm on each side.
\begin{eqnarray} \ln{(x-3)} + \ln{(x+4)} &=& 3\ln{2}\\ \ln{(x-3)(x+4)} &=& \ln{2}^3\\ (x-3)(x+4) &=& 8\\ x^2 + x-12 &=& 8\\ x^2+x-20 &=& 0\\ (x+5)(x-4) &=& 0\\ x = -5 \quad &or& \quad x = 4\\ \end{eqnarray}
As $x=-5$ is extraneous solution, we conclude that the solution for this problem is $x = 4$

Solve: $3^{(x-1)} = 29$

Solution:

Taking logarithm on both sides and simplifying to solve: \begin{eqnarray} 3^{(x-1)} &=& 29\\ \ln{3}^{(x-1)} &=& \ln{29}\\ (x-1)\ln{3} &=& \ln{29}\\ x-1 &=& \frac{\ln{29}}{\ln{3}}\\ x &=& \frac{\ln{29}}{\ln{3}} + 1\\ &=& \frac{\ln{29}+\ln{3}}{\ln{3}}\\ \end{eqnarray}
Using a calculator, we can approximate the solution as $x = 4.065$