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Precalculus : Properties of Logarithms

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Example Question #1 : Exponential And Logarithmic Functions

Solve for : $\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

Possible Answers:

$x = 4 \textrm{ or }x = 128$

$x = -7 \textrm{ or }x = -2$

$x = 1 \textrm{ or }x = \log_{2}7$

The correct solution set is not included among the other choices.

$x = 2 \textrm{ or }x = 7$

Correct answer:

The correct solution set is not included among the other choices.

Explanation:

$\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\log_{2} \left[ \left ( x+4 \right )\left ( x+5 \right ) \right ] = \log_{2} 6$

$2^{ \log_{2} \left[ \left ( x+4 \right )\left ( x+5 \right ) \right ] }=2^{ \log_{2} 6}$

(x + 4)(x + 5) = 6

FOIL: $x^{2} + 4x + 5x + 4 \cdot 5 = 6$

$x^{2} + 9x + 20 = 6$

$x^{2} + 9x + 20 - 6 = 6- 6$

$x^{2} + 9x + 14 = 0$

(x + 2) (x + 7) = 0

$x + 2 = 0 \textrm{ or }x + 7 = 0$

$x = -2 \textrm{ or } x = -7$

These are our possible solutions. However, we need to test them.

$\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\log_{2}\left ( -2+4 \right ) + \log_{2}\left ( -2+5 \right ) = \log_{2} 6$

$\log_{2}2 + \log_{2}3 = \log_{2} 6$

$\log_{2}2 = 1$

$\log_{2}3 = \frac{\ln 3}{\ln 2} \approx \frac{1.10}{0.69} \approx 1.59$

$\log_{2}6 = \frac{\ln 6}{\ln 2} \approx \frac{1.79}{0.69} \approx 2.59$

The equation becomes 1 + 1.59 = 2.59 . This is true, so x = -2 is a solution.

x = -7 :

$\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\log_{2}\left ( -7+4 \right ) + \log_{2}\left ( -7+5 \right ) = \log_{2} 6$

$\log_{2}\left (-3 \right ) + \log_{2}\left (-2 \right ) = \log_{2} 6$

However, negative numbers do not have logarithms, so this equation is meaningless. x = -7 is not a solution, and x = -2 is the one and only solution. Since this is not one of our choices, the correct response is "The correct solution set is not included among the other choices."

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Example Question #1 : Properties Of Logarithms

Solve for x: $\ln(x+2)-1=\ln(x-2)$

Possible Answers:

x=1

$x=\frac{e+2}{e-2}$

$x=\frac{2(e+1)}{e-1}$

$x=\frac{2(e-1)}{e+1}$

$x=\frac{e-2}{e+2}$

Correct answer:

$x=\frac{2(e+1)}{e-1}$

Explanation:

The key to simplifying this problem is by using the Natural Logarithm Quotient Rule $\ln(\frac{x}{y})= \ln x-\ln y$ .

Using algebraic manipulation to bring each natural logarithm to one side, we obtain:

$-1= \ln (x-2)- \ln (x+2)$

$1= \ln (x+2)- \ln (x-2)$

$1= \ln \frac{x+2}{x-2}$

$e = \frac{x+2}{x-2}$

ex-2e=x+2

ex-x=2e+2

x(e-1)=2(e+1)

$x=\frac{2(e+1)}{e-1}$

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Example Question #1 : Properties Of Logarithms

Solve for x: $\log_{3}x+\log_{3}(x-2)=1$

Possible Answers:

x=2 and x=-1

x=2

x=-1

x=3

x=3 and x=-1

Correct answer:

x=3

Explanation:

Using the logarithmic product rule $\log_b(xy)= \log_bx + \log_b y$ , we simplify as follows:

$\log_3(x^2-2x)=1$

x^2-2x=3

x^2-2x-3=0

Factoring this quadratic equation, we will obtain two roots.

(x-3)(x+1)=0

The roots are x=3 and x=-1 . However, the domain of the logarithmic function is $(0,\infty )$ . That is to say, it is not defined for numbers less than or equal to 0. So our final answer is

x=3

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Example Question #11 : Exponential And Logarithmic Functions

Simplify the expression as a single natural logarithm with a coefficient of one: $\ln4-2\ln2+\frac{1}{2}\ln25$ .

Possible Answers:

$-\ln12.5$

$\ln5$

$-\ln25$

$\ln12.5$

$\ln25$

Correct answer:

$\ln5$

Explanation:

Here we need to make use the power rule $\ln (x^y)= y\ln x$

$\ln4-2\ln2+\frac{1}{2}\ln25 = \ln4 - \ln(2^2) + \ln(25^\frac{1}{2})$

Recall that $x^\frac{1}{2}=\sqrt{x}$ , so we have

$\ln4 -\ln4+\ln\sqrt{25}$

$0+\ln5$

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Example Question #1 : Properties Of Logarithms

Evalute the equation $e^x+e^{-x}=2$ .

Possible Answers:

x=0

x=1

x=-2

x=2

$x=\ln2$

Correct answer:

x=0

Explanation:

We can rewrite $e^x+e^{-x}=2$ as $e^x+\frac{1}{e^{x}}=2$ , and then multiply each side by e^x . Doing so, we get

$e^{2x}+1=2e^x$

$e^{2x}-2e^{x}+1=0$

This is just a quadratic equation with $e^{x}$ replacing . Let us factor it just like a quadratic equation.

$(e^{x}-1)^2=0$

$e^{x}=1$

$x=\ln1$

x=0

x=0 in this case is a root with multiplicity of two, so there are two answers to this equality, both of them being .

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Example Question #5 : Properties Of Logarithms

Evalute the equation $5e^{x-1} -200=2600$ . Use a calculator to approximate the answer to three decimal places.

Possible Answers:

x=8.436

x=7.328

x=642.358

x=139.136

x=6.324

Correct answer:

x=7.328

Explanation:

In order to evaluate this equation, we have to do some algebraic manipulation first to get the exponential function isolated.

$5e^{x-1} -200=2600$

$5e^{x-1}=2800$

$e^{x-1}=560$

$x-1=\ln560$

$x=\ln560+1$

x=7.328

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Example Question #6 : Properties Of Logarithms

solve for x: $x=\log_42+\log_{\frac{1}{4}}2$ .

Possible Answers:

$x=\frac{1}{2}$

x=2

x=4

x=0

$x=\frac{1}{4}$

Correct answer:

x=0

Explanation:

Here we employ the use of the logarithm base change formula $\log_bx= \frac{\log_cx}{\log_cb}$ .

We could convert either $\log_42$ or $\log_{\frac{1}{4}}2$ to the other's base. Let's convert $\log_{\frac{1}{4}}2$ to a logarithm with base 4.

$\log_{\frac{1}{4}}2=\frac{\log_42}{\log_4\frac{1}{4}}$

Plugging this back in to the original equation,

$x=\log_42+\frac{\log_42}{\log_4\frac{1}{4}}$

$x=\frac{1}{2}+\frac{\frac{1}{2}}{-1}$

$x=\frac{1}{2}-\frac{1}{2}$

x=0

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Example Question #12 : Exponential And Logarithmic Functions

Calculators are not requried (and are strongly discouraged) for this problem.

Expand and simplify the following logarithm:

$\log_{4}\left ( 16x^2y^{-3}\right )$

Possible Answers:

$4 + 2\log_{4}x - 3\log_{4}y$

$2 + \frac{2}{3}\log_{4}\frac{x}{y}$

$2 + 2\log_{4}x - 3\log_{4}y$

$2 + 2\log_{4}x + \frac{1}{3}\log_{4}y$

$2 + 2\log_{4}x + 3\log_{4}y$

Correct answer:

$2 + 2\log_{4}x - 3\log_{4}y$

Explanation:

First expand the logarithm using the product property:

$\log_{b}(m*n) = \log_{b}m + \log_{b}n$

$\log_{4}\left ( 16x^2y^{-3}\right )$

$=\log_{4}\left ( 16*x^2*y^{-3}\right )$

$= \log_{4}16 + \log_{4}x^2 + \log_{4}y^{-3}$

We can evaluate the constant log on the left either by memorization, sight inspection, or deliberately re-writing 16 as a power of 4, which we will show here:

16 = 4^2 , so our expression becomes:

$= \log_{4}4^2 + \log_{4}x^2 + \log_{4}y^{-3}$

Now use the power property of logarithms:

$\log_{b}a^m = m\log_{b}a$

Rewrite the equation accordingly.

$= 2\log_{4}4 + 2\log_{4}x -3\log_{4}y^$

Note that the 3rd terms becomes negative because the exponent is negative. We will use one last log property to finish simplifying:

$\log_b{b}=1$

Accordingly,

$\log_44 = 1$ .

Now substitute and simplify:

$= 2(1) + 2\log_{4}x -3\log_{4}y^$

$= 2 + 2\log_{4}x -3\log_{4}y$

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Example Question #12 : Exponential And Logarithmic Functions

Simplify:

$\frac{-4\log_{10}x}{\frac{1}{2}\log_{10}9} - 2\log_3x$

Possible Answers:

$-8\log_3{\frac{1}{9x}}$

$-6\log_3x$

$-6\log_{10}{\frac{-8x}{9}}$

$-4\log_{10}{\frac{x}{9}}$

$\log_3x^{-6}$

Correct answer:

$\log_3x^{-6}$

Explanation:

First use the reversal of the logarithm power property to bring coefficients of the logs back inside the arguments:

$m * \log_{b}a = \log_ba^m$

Now apply this rule to every log in the formula and simplify:

$\frac{(-4)\log_{10}x}{\left ( \frac{1}{2} \right )\log_{10}9} + (- 2)\log_3x$

$\frac{\log_{10}x^{-4}}{\log_{10}9^{\frac{1}{2}}} + \log_3x^{-2}$

$\frac{\log_{10}x^{-4}}{\log_{10}3} + \log_3x^{-2}$

Next, use a reversal of the change-of-base theorem to collapse the quotient:

$\frac{\log_ma}{\log_mb} = \log_ba$

$\frac{\log_{10}x^{-4}}{\log_{10}3} = \log_3x^{-4}$

Substituting, we get:

$\log_{3}x^{-4} + \log_3x^{-2}$

Now combine the two using the reversal of the logarithm product property:

$\log_bm + \log_bn = \log_b (m*n)$

$\log_{3}x^{-4} + \log_3x^{-2} = \log_3\left ( x^{-4}*x^{-2}\right )$

$= \log_3x^{-6}$

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Example Question #131 : Pre Calculus

Find the inverse function of the following exponential function:

$y = \frac{2}{3}e^{4x+1}$

Possible Answers:

$y = \frac{1}{4}\ln {\frac{3}{2}x}-1$

$y = \frac{1}{4}\ln {\frac{2}{3}x}-\frac{1}{4}$

$y = \frac{1}{4}\ln {x}-\frac{1}{4}$

$y = \frac{1}{4}\ln {\frac{3}{2}x}-\frac{1}{4}$

$y = \frac{3}{8}\ln {x}-\frac{1}{4}$

Correct answer:

$y = \frac{1}{4}\ln {\frac{3}{2}x}-\frac{1}{4}$

Explanation:

Since we are looking for an inverse function, we start by swapping the x and y variables in our original equation.

$x = \frac{2}{3}e^{4y+1}$

Now we have to solve for y. To do this we have to work towards isolating y. First we remove the constant multiplier:

$\frac{3}{2}x = \frac{3}{2}*\frac{2}{3}e^{4y+1}$

$\frac{3}{2}x = e^{4y+1}$

Next we eliminate the base on the right side by taking the natural log of both sides.

$\ln\frac{3}{2}x = \ln e^{4y+1}$

$\ln\frac{3}{2}x = {4y+1}$

Subtract 1 and divide by 4:

$\ln\frac{3}{2}x - 1 = {4y+1} - 1$

$\ln\frac{3}{2}x - 1 = 4y$

$\frac{1}{4}\left ( \ln\frac{3}{2}x - 1 \right ) = 4y*\frac{1}{4}$

$\frac{1}{4}\ln {\frac{3}{2}x}-\frac{1}{4} = y$

$y = \frac{1}{4}\ln {\frac{3}{2}x}-\frac{1}{4}$

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Precalculus : Properties of Logarithms

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Exponential And Logarithmic Functions

Example Question #1 : Properties Of Logarithms

Example Question #1 : Properties Of Logarithms

Example Question #11 : Exponential And Logarithmic Functions

Example Question #1 : Properties Of Logarithms

Example Question #5 : Properties Of Logarithms

Example Question #6 : Properties Of Logarithms

Example Question #12 : Exponential And Logarithmic Functions

Example Question #12 : Exponential And Logarithmic Functions

Example Question #131 : Pre Calculus

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