Exponents and Logarithms - SAT Subject Test in Math I

Card 0 of 17

Question

Rewrite as a single logarithmic expression:

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

Answer

Using the properties of logarithms

$n \ln a = \ln a^{n}$ and $\ln a - \ln b = \ln \frac{a}{b}$ ,

we simplify as follows:

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

$= \ln x - \ln (x + 2)^{2}$

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

$= \ln \frac{x}{x^{2} +4x + 4}$

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Question

How many elements are in a set that has exactly 128 subsets?

Answer

A set with elements has $2 ^{N}$ subsets.

Solve:

$2 ^{N} = 128$

$\ln 2 ^{N} = \ln 128$

$N \ln 2 = \ln 128$

$N = \frac{\ln 128}{\ln 2} = \frac{4.8520 }{0.6931} = 7$

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Question

Solve: $log_{3}27$

Answer

In order to solve this problem, covert 27 to the correct base and power.

$log_{3}27 = log_{3}: 3^3$

Since , the correct answer is .

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Question

Simplify $(x^{-4})^7$

Answer

When an exponent is raised by another exponent, we just multiply the powers.

$(x^{-4})^7=x^{-4*7}=x^{-28}$

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Question

Simplify:

Answer

When adding exponents, we don't add the exponents or multiply out the bases. Our goal is to see if we can factor anything. We do see three . Let's factor.

$3^9+3^9+3^9=3^9(1+1+1)=3^9(3)=3^{10}$ Remember when multiplying exponents, we just add the powers.

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Question

Solve and simplify.

$\sqrt[3]{125}$

Answer

$\sqrt[3]{125}$ Another way to write this is . The only number that makes is .

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Question

Simplify:

$\log_3729$

Answer

$\log_3729$ is the same as . Let's factor out . It's the same as . Therefore which is the answer to our question.

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Question

Simplify:

$\log50-\log5$

Answer

When dealing with subtraction in regards to logarithms, it's the same as dividing the numbers.

$\log50-\log5=\frac{\log50}{\log5}=\log \frac{50}{5}=\log10=1$

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Question

Simplify:

$\log_618+\log_672$

Answer

When dealing with addition in regards to logarithms, it's the same as multiplying the numbers.

$\log_618+\log_672=\log_6(18*72)=\log_61296=4$

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Question

Solve: $(x^{2})^{5}$ when .

Answer

Power rule says when an exponent is raised to another exponent, you must multiply the exponents.

So and our expression is now $x^{10}$ .

Plug in the given value to get

$2^{10}=1024$ .

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Question

Simplify $log_{10}(\frac{10}{100})$

Answer

One of the properties of log is that $\small log_{x}(\frac{n}{m})=log_{x}(n)-log_{x}(m)$

Applying that principle to this problem:

$log_{10}(\frac{10}{100})=log_{10}(10)-log_{10}(100)$

Simplifying the log base 10

$log_{10}(10)=log_{10}(10^1)=1$

$log_{10}(100)=log_{10}(10^2)=2$

Plug in the values to the first equation:

$log_{10}(\frac{10}{100})=log_{10}(10)-log_{10}(100)=1-2=-1$

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Question

Give the set of real solutions to the equation

$2 ^{2x+2} + 11 \cdot 2^{x} - 3 = 0$

(round to the nearest hundredth, if applicable)

Answer

Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first term:

$2 ^{2x+2} + 11 \cdot 2^{x} - 3 = 0$

$2 ^{2x } \cdot 2 ^{2}+ 11 \cdot 2^{x} - 3 = 0$

$4 \cdot( 2 ^{x } ) ^{2}+ 11 \cdot 2^{x} - 3 = 0$

Substitute for $2^{x}$ ; the equation becomes

$4t ^{2}+ 11t- 3 = 0$ ,

which is quadratic in terms of . The trinomial might be factorable using the method, where we split the middle term with integers whose product is $4 \cdot (-3) = -12$ and whose sum is 11. By trial and error, we find the integers to be 12 and , so the equation can be written as follows:

$4t ^{2} -t + 12t- 3 = 0$

Factoring by grouping:

$(4t ^{2} -t )+( 12t- 3 )= 0$

By the Zero Product Rule, one of these two factors must be equal to 0.

If , then .

Substituting $2^{x}$ back for , we get

$2^{x}= -3$ .

This is impossible, since any power of a positive number must be positive.

If , then:

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

Substituting $2^{x}$ back for , we get

$2^{x}= \frac{1}{4}$

Since $\frac{1}{4} = \frac{1}{2^{2}} = 2^{-2}$ ,

it holds that $2^{x}= 2^{-2}$ , and , the only solution.

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Question

Give the set of real solutions to the equation

$3 ^{2x+1} -28 \cdot 3^{x}+9= 0$

(round to the nearest hundredth, if applicable)

Answer

Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first term:

$3 ^{2x+1} -28 \cdot 3^{x}+9= 0$

$3 ^{2x } \cdot 3^{1 } -28 \cdot 3^{x}+9= 0$

$3 \cdot 3 ^{2x } -28 \cdot 3^{x}+9= 0$

$3 \cdot (3 ^{x } )^{2 } -28 \cdot 3^{x}+9= 0$

Substitute for $3^{x}$ ; the equation becomes

$3t^{2 } -28t+9= 0$

which is quadratic in terms of . The trinomial might be factorable using the method, where we split the middle term with integers whose product is $3 \cdot 9 = 27$ and whose sum is . By trial and error, we find the integers to be and , so the equation can be rewritten as follows:

$3t^{2 } -27t -t +9= 0$

Factoring by grouping:

$(3t^{2 } -27t )-( t -9)= 0$

By the Zero Product Rule, one of these two factors must be equal to 0.

If , then .

Since $3 ^{2} = 9$ , then substituting this as well as substituting $3^{x}$ back for , we get

$3^{x} = 3 ^{2}$ ,

and

If , then

$t = \frac{1}{3}$

Since $3 ^{-1} = \frac{1}{3}$ , then substituting this as well as substituting $3^{x}$ back for , we get

$3^{x} = 3 ^{-1}$ , and

The solution set is therefore $\left { -1, 2 \right }$

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Question

Solve for (round to the nearest hundredth):

$3 ^{x+ 4} = 2^{x+ 9}$

Answer

Take the natural logarithm of both sides:

$3 ^{x+ 4} = 2^{x+ 9}$

$\ln 3 ^{x+ 4} = \ln 2^{x+ 9}$

By Logarithm of a Power Rule, the above becomes

$(x+ 4) \ln 3 = (x+ 9) \ln 2$

After distributing, solve for :

$x \ln 3 + 4 \ln 3 = x \ln 2+ 9 \ln 2$

$x \ln 3 + 4 \ln 3 - x \ln 2 - 4 \ln 3 = x \ln 2+ 9 \ln 2 - x \ln 2 - 4 \ln 3$

$x \ln 3 - x \ln 2 = 9 \ln 2 - 4 \ln 3$

Factor out the left side, then divide:

$x( \ln 3 - \ln 2) = 9 \ln 2 - 4 \ln 3$

$\frac{x( \ln 3 - \ln 2) }{\ln 3 - \ln 2}=\frac{ 9 \ln 2 - 4 \ln 3}{\ln 3 - \ln 2}$

$x=\frac{ 9 \ln 2 - 4 \ln 3}{\ln 3 - \ln 2}$

Substituting the values of the logarithms:

$x \approx \frac{ 9 (0.6931) - 4 (1.0986)}{1.0986 - 0.6931}$

$x\approx \frac{ 1.8439}{0.4055}$

$x \approx 0.45472$

This rounds to 0.45.

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Question

Solve for :

$2^{x} = 8 ^{2-x}$

Answer

$8 = 2^{3}$ , so the equation

$2^{x} = 8 ^{2-x}$

can be rewritten as:

$2^{x} =( 2 ^{3} ) ^{2-x}$

By the Power of a Power rule:

$2^{x} = 2 ^{3 (2-x)}$

It follows that

Solving for :

$x = 3 \cdot 2- 3 \cdot x$

$4x \div 4 = 6 \div 4$

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

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Question

$\textup{Solve for }x:$

$2 ^{2x} +4 = 2^{x+2}$

Answer

By the Power of a Power and Product of Power Rules, we can rewrite this equation as

$2 ^{2x} +4 = 2^{x+2}$

$(2 ^{ x }) ^{2} +4 = 2^{2} \cdot 2^{x}$

$(2 ^{ x }) ^{2} +4 = 4 \cdot 2^{x}$

Substitute for $2^{x}$ ; the resulting equation is the quadratic equation

$t^{2} +4 = 4t$ ,

which can be written in standard form by subtracting from both sides:

$t^{2} +4 - 4t = 4t - 4t$

$t^{2}- 4t +4 = 0$

The quadratic trinomial fits the perfect square trinomial pattern:

$t^{2}- 2 \cdot t \cdot 2 + 2^{2} = 0$

$(t-2)^{2} = 0$

By the square root principle,

Substituting $2^{x}$ for :

$2^{x} = 2$

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

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Question

Evaluate: $xe^{ln(4-2x)}$

Answer

An exponential base raised to the natural log will eliminate, leaving only the terms of the power. This is a log rule that can be used to simplify the expression.

$xe^{ln(4-2x)} = x(4-2x)$

Distribute the x variable through the binomial.

The answer is:

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