Properties of Logarithms - Pre-Calculus

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Question

Simplify

$\small \log ,3,,+,,\log,7$

Answer

By the property of the addition of logarithms with the same base

$\small \small \log,a,,+,,\log,b=\log,ab$

As such

$\small \small \log ,3,,+,,\log,7=\log,3\cdot7=\log,21$

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Question

Evaluate a logarithm.

What is $log_{10}1000$ ?

Answer

The derifintion of logarithm is:

$log_{a}y=x \Leftrightarrow a^{x}=y.$

In this problem,

$10^{3}=1000$

Therefore,

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Question

Given the equation $ln(e^{3x})=15$ , what is the value of ? Use the inverse property to aid in solving.

Answer

The natural logarithm and natural exponent are inverses of each other. Taking the of will simply result in the argument of the exponent.

That is $ln(e^{3x})=3x$

Now, , so

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Question

What is equivalent to?

Answer

Using the properties of logarithms,

$\log(a)-\log(b)=\log \bigg( \frac{a}{b}\bigg)$

the expression can be rewritten as

$\ln(2x)-\ln(2)=\ln\bigg(\frac{2x}{2}\bigg)$

which simplifies to $\ln(x)$ .

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Question

Condense the following equation:

Answer

Let's use the properties of logarithms to condense this equation. We will use the follwing three properties

Power Rule $b\cdot log(x)=log(x^b)$

Product Rule

Quotient Rule $log(x)-log(y)=log(\frac{x}y{})$

Let's first use the power rule to rewrite the second term

Then, we'll use the product rule to combine the first threeterms

Lastly, we'll use the quotient rule to combine into one term

$log(\frac{3xy^4}{z})$

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Question

Solve for in the following logarithmic equation:

$log_{10} x^2 - log_{10} 2x = 2$

Answer

Using the rules of logarithms,

$\log a - \log b = \log \bigg(\frac{a}{b}\bigg)$

Hence,

$\log x^2 - \log 2x = log \bigg(\frac{x^2}{2x}\bigg) = log \bigg(\frac{x}{2}\bigg) = 2$

So exponentiate both sides with a base 10:

$10^{log_{10} \frac{x}{2}} = 10^{2}$

The exponent and the logarithm cancel out, leaving:

$\frac{x}{2}=10^2$

$x=100\cdot 2=200$

This answer does not match any of the answer choices, therefore the answer is 'None of the other choices'.

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Question

Express the log in its expanded form:

$\log\bigg(\frac{x^3y}{z}\bigg)$

Answer

You need to know the Laws of Logarithms in order to solve this problem. The ones specifically used in this problem are the following:

$\log(xy) = \log (x)+\log(y)$

$\log\bigg(\frac{x}{y}\bigg) = \log(x)-\log(y)$

$\log(x^y) = y\cdot \log(x)$

Let's take this one variable at a time starting with expanding z:

$\log\bigg(\frac{x^3y}{z}\bigg) = \log(x^3y)-\log(z)$

Now y:

$\log(x^3y)-\log(z) = \log(x^3)+\log(y)-\log(z)$

And finally expand x:

$\log(x^3)+\log(y)-\log(z) = 3\log(x)+\log(y)-\log(z)$

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Question

Find the value of the sum of logarithms by condensing the expression.

$\ln(2)+\ln\bigg(\frac{1}{2}\bigg)$

Answer

By the property of the sum of logarithms,

$\ln(2)+\ln\bigg(\frac{1}{2}\bigg)=\ln\bigg(2\cdot\frac{1}{2}\bigg)=\ln(1)=0$ .

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Question

Solve the following logarithmic equation:

$2\log _3(x)=\log _3(4)+\log _3(x-1)$

Answer

In order to solve this equation, we must apply several properties of logarithms. First we notice the term on the left side of the equation, which we can rewrite using the following property:

$a\log_b (x)=\log_b (x^a)$

Where a is the coefficient of the logarithm and b is some arbitrary base. Next we look at the right side of the equation, which we can rewrite using the following property for the addition of logarithms:

$\log _ba+\log _bc=\log_b (a\cdot c)$

Using both of these properties, we can rewrite the logarithmic equation as follows:

$2\log _3(x)=\log _3(4)+\log _3(x-1)$

$\log _3(x^2)=\log _3(4(x-1))$

We have the same value for the base of the logarithm on each side, so the equation then simplifies to the following:

$x^2=4x-4\rightarrow x^2-4x+4=0$

Which we can then factor to solve for :

$(x-2)(x-2)=0\rightarrow x=2$

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Question

Expand the following logarithmic expression:

$\log \left(\frac{8x^2y}{z^3}\right)$

Answer

We start expanding our logarithm by using the following property:

$\log \left(\frac{a}{b}\right)=\log(a)-\log (b)$

$\log \left(\frac{8x^2y}{z^3}\right)=\log (8x^2y)-\log (z^3)$

Now we have two terms, and we can further expand the first term with the following property:

$\log (abc)=\log (a)+\log (b)+\log (c)$

$\log (8x^2y)=\log (8)+\log (x^2)+\log (y)$

$\log \left(\frac{8x^2y}{z^3}\right)=\log (8)+\log (x^2)+\log (y)-\log (z^3)$

Now we only have two logarithms left with nonlinear terms, which we can expand using one final property:

$\log (a^b)=b\log (a)$

Using this property on our two terms with exponents, we obtain the final expanded expression:

$\log \left(\frac{8x^2y}{z^3}\right)=\log (8)+2\log (x)+\log (y)-3\log (z)$

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Question

Condense the following logarithmic equation:

$3\log (y)+\log (4)+\log (x)-2\log (z)$

Answer

We start condensing our expression using the following property, which allows us to express the coefficients of two of our terms as exponents:

$b\log (a)=\log (a^b)$

$3\log (y)+\log (4)+\log (x)-2\log (z)$

$\log(y^3)+\log(4)+\log(x)-\log(z^2)$

Our next step is to use the following property to combine our first three terms:

$\log (a)+\log(b)+\log(c)=\log(abc)$

$\log(y^3)+\log(4)+\log(x)-\log(z^2)$

$\log(4xy^3)-\log(z^2)$

Finally, we can use the following property regarding subtraction of logarithms to obtain the condensed expression:

$\log (a)-\log (b)=\log (\frac{a}{b})$

$\log(4xy^3)-\log(z^2)$

$\log \left(\frac{4xy^3}{z^2}\right)$

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Question

What is another way of writing

$\frac{\ln(x^b)}{\ln(y^a)}$ ?

Answer

The correct answer is

$\frac{\ln(x^{1/a})}{\ln(y^{1/b})}$

Properties of logarithms allow us to rewrite $\ln y^a$ and $\ln x^b$ as $\ln y^a=a\ln y$ and $b\ln x$ , respectively. So we have

$\frac{\ln(x^b)}{\ln(y^a)}=\frac{b}{a}\frac{\ln(x)}{\ln(y)}=\frac{1/a}{1/b}\frac{\ln(x)}{\ln(y)}$

Again, we use the logarithm property

$\frac{1}{a}\ln x=\ln x^{1/a}$

to get

$\frac{\ln(x^b)}{\ln(y^a)}=\frac{1/a}{1/b}\frac{\ln(x)}{\ln(y)}=\frac{\ln(x^{1/a})}{\ln(y^{1/b})}$

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Question

Solve the equation for .

$\ln\frac{cx}{ab+c} = b+c$

Answer

We solve the equation as follows:

$\ln \frac{cx}{ab+c} = b+c$

Exponentiate both sides.

$\frac{cx}{ab+c}= e^{b+c}$

Apply the power rule on the right hand side.

$\frac{cx}{ab+c} = e^be^c$

Multiply by .

Divide by .

$x =\frac{e^be^c(ab+c)}{c}$

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Question

Expand

$\log_{2}\left(\frac{2}{9x+2}\right)$ .

Answer

To expand

$\log_{2}\left(\frac{2}{9x+2}\right)$ , use the quotient property of logs.

The quotient property states:

$\log_{b} (xy)=\log_{b}x-\log_{b}y$

Substituting in our given information we get:

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

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Question

Condense the logarithm

Answer

In order to condense the logarithmic expression, we use the following properties

As such

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Question

Express $log\left(\frac{xy^2}{h}\right)$ in its expanded, simplified form.

Answer

Using the properties of logarithms, expand the logrithm one step at a time:

When expanding logarithms, division becomes subtration, multiplication becomes division, and exponents become coefficients.

$log\left(\frac{xy^2}{h}\right)=log(xy^2)-log(h)$ .

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Question

Which of the following correctly expresses the following logarithm in expanded form?

$log\left(\frac{5x^2}{y^3}\right)$

Answer

Begin by recalling a few logarithm rules:

When adding logarithms of like base, multiply the inside.

When subtracting logarithms of like base, subtract the inside.

When multiplying a logarithm by some number, raise the inside to that power.

Keep these rules in mind as we work backward to solve this problem:

$log\left(\frac{5x^2}{y^3}\right)$

Using rule 2), we can get the following:

Next, use rule 1) on the first part to get:

Finally, use rule 3) on the second and third parts to get our final answer:

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Question

Solve for :

Answer

First, simplify the logarithmic expressions on the left side of the equation:

can be re-written as .

Now we have:

.

The left can be consolidated into one log expression using the subtraction rule:

$log\left(\frac{27}{x^2}\right)=log(x)$ .

We now have log on both sides, so we can be confident that whatever is inside these functions is equal:

$\frac{27}{x^2}=x$ to continue solving, multiply by on both sides:

take the cube root:

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Question

$(\ln (x^2))^2=\ln (x^4)+3$ .

Solve for .

Answer

First bring the inside exponent in front of the natural log.

$(\ln (x^2))^2=\ln (x^4)+3\rightarrow(2\ln (x))^2=4\ln (x)+3$ .

Next simplify the first term and bring all the terms on one side of the equation.

$4(\ln (x))^2-4\ln (x)-3=0$ .

Next, let set

$\ln (x)=A$ , so .

Now use the quadratic formula to solve for .

$\ \frac{-b\pm \sqrt{b^2-4(a)(c)}}{2a}\ \=\frac{4\pm\sqrt{16-4(4)(-3)}}{2(4)}\ \=\frac{4\pm \sqrt{16+48}}{8}\ \=4\pm \frac{\sqrt{64}}{8}\ \=\frac{4\pm 8}{8}\ \=\frac{3}{2}; -\frac{1}{2}$

and thus, $A=-\frac{1}{2}$ and $A=\frac{3}{2}$ .

Now substitute with $\ln (x)$ .

So, $\ln (x)=-\frac{1}{2}\rightarrow \O$ since and $\ln (x)=\frac{3}{2}\rightarrow x=e^{3/2}$ .

Thus, $x=e^{3/2}$ .

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Question

Write the expression in the most condensed form: $4\ln(x)-\frac{1}{2}\ln(x+4)$

Answer

$4\ln(x)-\frac{1}{2}\ln(x+4)$

Use the Power property of Logarithms:

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

Rewrite the fractional exponent:

$\ln(x)^4-\ln\sqrt{(x+4)}$

Condense into a fraction using the Quotient property of Logarithms:

$\mathbf{\ln\frac{x^4}{\sqrt{x+4}}}$

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