Practice questions

First, rewrite the equation so that both the exponents have same base:

$(4^2)^{2x+1}=(4^3)^{x+3}$

$(4)^{2(2x+1)}=(4)^{3(x+3)}$

Now, the bases are the same, so the exponents should be the same:

$2(2x+1)=3(x+3)$

Now, solve for $x$:

$4x+2=3x+9$

Solving for $x$:

$x=7$

Write the logarithmic equation in exponential form:

$4^3=2x+4$

$64=2x+4$

Solving for $x$:

$x=30$

You cannot write this equation with same base on both sides. So take log on both sides:

$\log(2^{x+3})=\log(5^x)$

$(x+3)\log(2)=x\log(5)$

$x\log(2)+3\log(2)=x\log(5)$

Bring the $x$ terms together:

$x\log(2)-x\log(5)=-3\log(2)$

Factor out the $x$:

$x(\log(2)-\log(5))=-3\log(2)$

Divide $(\log(2)-\log(5))$ to get $x$:

$x=\dfrac{-3\log(2)}{\log(2)-\log(5)}$

After using the calculator:

$x=2.269$

First combine the two logarithms on the left:

$\log_2(x(x-7))=3$

Now, write in exponential form:

$2^3=x(x-7)$

$8=x^2-7x$

This is a quadratic equation, write in standard form to solve:

$0=x^2-7x-8$

i.e., $x^2-7x-8=0$

Factoring:

$(x-8)(x+1)=0$

Applying zero-product rule:

$x-8=0$ or $x+1=0$

$x=8$ or $x=-1$

$x=-1$ makes a negative number inside the logarithm, so this cannot be a solution.

So, the solution is

$x=8$

Write the coefficients of log as exponents using the formula $\log_a{x^p}=p\log x$

$=\log_p(x^3)+\log_p(y^{1/2})-\log_p(z^{3/2})$

Combine them:

$=\log_p\left(\dfrac{x^3y^{1/2}}{z^{3/2}}\right)$

You can also write this as

$=\log_p\left(\dfrac{x^3\sqrt{y}}{\sqrt{z^3}}\right)$

Use the change of base formula. Change the base to $e$ as there is an $e$ in the log. (You can also change the base to 10).

$\log_\pi(e)=\dfrac{\ln(e)}{\ln (\pi)}$

Since, $\ln (e)=1$:

 $=\dfrac{1}{\ln (\pi)}$

 $=\dfrac{1}{1.14473}$

 $=0.8736$, after rounding to the four decimal places.

Use the change of base formula:

$3\log_6 2.75= 3\cdot \dfrac{\log(2.75)}{\log(6)}$

Use the calculator:

  $= 3\cdot \dfrac{0.4393327}{0.778151}$

  $= 3\cdot 0.564585$

  $= 1.6938$

Graph of $y=\left(\dfrac{2}{3}\right )^x$:

$x$	$y$	$(x, y)$
$-2$
$-1$
$0$
$1$
$2$

Logarithmic function is the inverse of the exponential function.

So, to draw the graph of $y=\log_2{x}$, draw the graph of $y=2^x$. Then draw a diagonal line as a mirror. The logarithmic function is the mirror image.

For the interest compounded continuously, you should use the following formula:

$A=P e^{rt}$

$t$ is the number of years, $A$ is the amount after $t$ years, $P$ is the Principal (initial deposit or loan if money is borrowed) and $r$ is the interest rate.

You are given:

$P=4000\:\$$; $r=3.5\%=3.5/100=0.035$; and $t=6$ years. You need to find the amount $A$ after $6$ years. Substituting, $P$,$r$ and $t$ in the equation:

$A=4000 e^{0.035\cdot 6}$

$A=4000 e^{0.035\cdot 6}$

$ =4000 e^{0.21}$

$ =4000 \cdot 1.2337$

$ =4934.71 \:\$$

This is one of the answer.

Next, you need to find how long does it take to double the amount. That is you need to find the time, $t$ to double the amount.

Amount is doubled, so, $A=2P$. Substituting this, $A$ and $r$ in the formula:

$2P=Pe^{0.035t}$

Divide out $P$, and solve for $t$:

$2=e^{0.035t}$

To solve, take $\ln$ on both sides:

$\ln (2)= \ln(e^{0.035t})$

$\ln (2)= 0.035t\,\ln(e)$

Putting, $\ln(e)=1$:

$\ln (2)= 0.035t$

Divide $0.035$ to get $t$:

$t=\ln(2)/0.035=19.8$ years

The equation for the amount after $t$ years is given

$A(t)=29,000(1.03)^t$.

Initial deposit amount and the interest rate are included in the equation, so don't worry about those numbers.

You need to find the time $t$ to reach the amount, $35,000$.

Substitute, $A=35,000$ and solve for $t$:

$35000=29000(1.03)^t$

Divide $29000$ on both sides to isolate the exponent:

$1.2069=1.03^t$

This is an exponential equation. To solve for $t$, take log on both sides:

$\log(1.2069)=\log(1.03^t)$

$\log(1.2069)=t\,\log(1.03)$

Divide, $\log(1.03)$ on both sides to get $t$:

$\dfrac{\log(1.2069)}{\log(1.03)}=t$

Using the calculator:

$t=6.36$

This is the answer. So, after $6.36$ years, the loan amount will reach \$35,000.