Exponents and Logarithms - SAT Math

Card 0 of 180

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

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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Question

Solve for :

$5^{x} = 432$

Give the solution to the nearest hundredth.

Answer

One way is to take the common logarithm of both sides and solve:

$5^{x} = 432$

$\log \left (5^{x} \right )= \log 432$

$x \log 5= \log 432$

$x = \frac{\log 432}{\log 5} \approx \frac{2.6355}{0.6990} \approx 3.77$

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Question

Solve for :

$\left (e+1 \right ) ^{k} = 100$

Give your answer to the nearest hundredth.

Answer

Take the common logarithm of both sides and solve for :

$\left (e+1 \right ) ^{k} = 100$

$\log \left (e+1 \right ) ^{k} = \log 100$

$k \log \left (e+1 \right ) = 2$

$k= \frac{2}{ \log \left (e+1 \right ) }$

$k= \frac{2}{ \log \left (2.7183+1 \right ) } \approx \frac{2}{ \log 3.7183 } \approx \frac{2}{1.3133} \approx 3.51$

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Question

Solve for :

$10^{t} + 1 = e$

Give your answer to the nearest hundredth.

Answer

$10^{t} + 1 = e$

$10^{t} = e - 1$

Take the common logarithm of both sides and solve for :

$\log 10^{t} = \log \left (e - 1 \right )$

$t = \log (e - 1) \approx \log (2.718-1) \approx \log 1.718 \approx 0.24$

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Question

Solve for :

$e^{x} = \pi$

Give your answer to the nearest hundredth.

Answer

Take the natural logarithm of both sides and solve for :

$e^{x} = \pi$

$\ln e^{x} = \ln \pi$

$x = \ln \pi \approx \ln 3.1416 \approx 1.14$

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Question

To the nearest hundredth, solve for :

$2^{x+4} = 5^{x}$

Answer

Take the common logarithm of both sides, then solve the resulting linear equation.

$2^{x+4} = 5^{x}$

$\log 2^{x+4} = \log 5^{x}$

$(x + 4) \log 2 = x \log 5$

$x\log 2 + 4\log 2 = x \log 5$

$4\log 2 = x \log 5 - x\log 2$

$4\log 2 = x\left ( \log 5 - \log 2 \right )$

$x = \frac{4\log 2 }{ \log 5 - \log 2 }$

$x \approx \frac{4 \cdot 0.3010}{0.6990-0.3010} \approx 3.03$

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