Exponents and Logarithms - SAT Subject Test in Math II

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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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Question

To the nearest hundredth, solve for :

$2^{x+10}= 10^{x+2}$

Answer

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

$2^{x+10}= 10^{x+2}$

$\log 2^{x+10}= \log 10^{x+2}$

$\left (x+10 \right )\log 2 = x+2$

$x\log 2 +10 \log 2 = x+2$

$x\log 2 +10 \log 2 - 2 - x \log 2= x+2 - 2 - x \log 2$

$-2 +10 \log 2 = x - x \log 2$

$x \left (1 - \log 2 \right )= -2 +10 \log 2$

$x =\frac{ -2 +10 \log 2}{1 - \log 2 }$

$x \approx \frac{ -2 +10 \cdot 0.3010}{1 - 0.3010 } \approx 1.45$

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Question

Solve for :

$100 ^{x}= 66$

Answer

Take the common logarithm of both sides:

$100 ^{x}= 66$

$\log 100 ^{x}= \log 66$

$x \log 100 = \log 66$

$x \cdot 2 = \log 66$

$x = \frac{ \log 66}{2}$

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Question

Solve for :

$3 + \log 4 + \log 6 = \log N$

Answer

The base of the common logarithm is 10, so

$3 = \log 10^{3} = \log 1,000$

The sum of three logarithms is the logarithm of the product of the three powers, so:

$3 + \log 4 + \log 6$

$=\log 1,000 + \log 4 + \log 6$

$= \log (1,000 \times 4 \times 6) = \log 24,000$

Therefore,

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Question

Solve for :

$\log 20 - 2 + \log 6 = \log N$

Answer

The base of the common logarithm is 10, so

$2 = \log 10^{2} = \log 1 00$

The sum of logarithms is the logarithm of the product of the three powers, and the difference of logarithms is the logarithm of the quotient of their powers. Therefore,

$\log 20 - 2 + \log 6$

$= \log 20 - \log 100 + \log 6$

$= \log \frac{20 \times 6 }{100} = \log \frac{120 }{100} =\log \frac{6}{5}$

$N = \frac{6}{5} = 1 \frac{1}{5}$

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Question

To the nearest hundredth, solve for :

$5^{N} = 10^{4-N}$

Answer

$5^{N} = 10^{4-N}$

$\log 5^{N} = \log 10^{4-N}$

$N \log 5 = 4 - N$

$N \log 5 + N = 4$

$N \left (1+ \log 5 \right ) = 4$

$N = \frac{4}{1+ \log 5 } \approx \frac{4}{1+ 0.6990 } \approx 2.35$

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Question

Give the set of real solutions to the equation

$2^{2x+ 4} - 2 ^{x+4} + 3 = 0$

(round to the nearest hundredth, if applicable)

Answer

Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first two terms strategically:

$2^{2( x+2)} - 2 ^{(x+2)+ 2 } + 3 = 0$

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

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

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

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

Factor this as

$(t+ \square)(t+ \square) = 0$

by finding two integers whose product is 3 and whose sum is . Through some trial and error we find , so we can write

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

If , then .

Substituting $2^{x+2 }$ back for :

$2^{x+2 } = 1$

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

.

If , then .

Substituting $2^{x+2 }$ back for :

$2^{x+2 } = 3$

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

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

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

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

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

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

$x \approx \frac{1.0986 - 2 (0.6931 )}{ 0.6931 }$

$x \approx -0.42$

Both can be confirmed to be solutions by substitution.

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Question

Give the solution set:

$6^{2x+1} + 6 = 6 ^{x+2} + 6 ^{x}$

Answer

Rewrite by taking advantage of the Product of Powers Property and the Power of a Power Property:

$6^{2x+1} + 6 = 6 ^{x+2} + 6 ^{x}$

$6^{1 } \cdot 6^{2x} + 6 = 6 ^{2} \cdot 6 ^{x } + 6 ^{x}$

$6 \cdot 6^{2x} + 6 =36 \cdot 6 ^{x } + 6 ^{x}$

$6 \cdot (6^{x } )^{2 }+ 6 = 37 \cdot( 6 ^{x })$

Substitute for $6^{x}$ ; the resulting equation is the quadratic equation

$6 t ^{2 }+ 6 = 37t$

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

$6 t ^{2 }+ 6 - 37t = 37t - 37t$

$6 t ^{2 }- 37t + 6 = 0$

The trinomial can be factored by the method, Look for two integers with sum and product $6 \cdot 6 = 36$ ; by trial and error, we find they are , so the equation can be rewritten and solved by grouping:

$6 t ^{2 }- 36t - t + 6 = 0$

$(6 t ^{2 }- 36t) - (t - 6) = 0$

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

Either

$\frac{6t}{6} = \frac{1}{6}$

$t = \frac{1}{6}$

Substituting $6^{x}$ back for :

$6^{x} = \frac{1}{6}$

$6^{x} = 6^{-1 }$

Or:

Substituting $6^{x}$ back for :

$6^{x} = 6$

$6^{x} = 6^{1}$

The solution set, as can be confirmed by substituting in the equation, is $\left { -1, 1 \right }$ .

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Question

Solve for :

$9 ^{x+6} = 27 ^{2x}$

Answer

$9 = 3^{2}$ and $27 = 3 ^{3}$ , so,

$9 ^{x+6} = 27 ^{2x}$

can be rewritten as

$( 3 ^{2}) ^{x+6} = (3 ^{3})^{2x}$

Applying the Power of a Power Rule,

$3 ^{2 \cdot (x+6)} = 3 ^{3 \cdot 2x}$

$3 ^{2 x+12} = 3 ^{6x}$

$\frac{12 }{4}= \frac{4x}{4}$

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Question

To the nearest hundredth, solve for : $7^{2x} = 5 ^{x}$ .

Answer

Take the natural logarithm of both sides:

$7^{2x} = 5 ^{x}$

$\ln 7^{2x} = \ln 5 ^{x}$

By the Logarithm of a Power Rule the above becomes

$2x \ln 7 = x \ln 5$

Solve for :

$2x \ln 7 - x \ln 5 = x \ln 5 - x \ln 5$

$(2 \ln 7 - \ln 5) x = 0$

$\frac{(2 \ln 7 - \ln 5) x }{2 \ln 7 - \ln 5}= \frac{0 }{2 \ln 7 - \ln 5}$

.

This is not among the choices given.

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Question

Solve the equation: $3^9 = 9^{3x-3}$

Answer

Rewrite the base of the right side.

$3^9 = 3^{2(3x-3)}$

Simplify the right side.

Add 6 on both sides.

Divide by 6 on both sides.

$\frac{15}{6} =\frac{ 6x}{6}$

The answer is: $\frac{5}{2}$

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Question

Solve: $8^3 = 2^{2x}$

Answer

Change the base of the left side to base two.

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

The equation becomes:

$2^9 = 2^{2x}$

Set the exponents equal since they have similar bases.

Divide by 2 on both sides.

$\frac{9}{2}=\frac{2x}{2}$

The answer is: $\frac{9}{2}$

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Question

Simplify:

$\frac{(a^3b^2c^7)^{-3}}{(a^7b^4c^{-2})^2}$

Answer

Start by distributing the exponent in both the numerator and the denominator. Recall that when an exponent is raised to the exponent, you will need to multiply the two numbers together.

$\frac{(a^3b^2c^7)^{-3}}{(a^7b^4c^{-2})^2}=\frac{a^{-9}b^{-6}c^{-21}}{a^{14}b^8c^{-4}}$

Next, recall that when you have numbers with different exponents, but the same base, subtract the exponent found in the denominator from the exponent in the numerator.

$\frac{a^{-9}b^{-6}c^{-21}}{a^{14}b^8c^{-4}}=a^{-23}b^{-14}c^{-17}$

Recall that you can flip the fraction to make the exponents positive.

$a^{-23}b^{-14}c^{-17}=\frac{1}{a^{23}b^{14}c^{17}}$

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Question

Solve $(\frac{1}{36})^{x+3}=\frac{1}{216}$ .

Answer

The first thing we need to do is find a common base. This can be tricky to do, but guessing and checking a little shows that:

$(\frac{1}{216})=(\frac{1}{36})^{\frac{3}{2}}$

Plugging that back in to the original equation:

$(\frac{1}{36})^{x+3}=(\frac{1}{36})^{\frac{3}{2}}$

Now that our bases are the same, we can cancel them:

$x+3=\frac{3}{2}$

From here, it's much easier to solve using simple algebra:

$x=-\frac{3}{2}$

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Question

Solve

$-9e^{x + 1} = 47$

Answer

The first thing we need to do is find a common base. However, because one of the bases has an in it (an irrational number), and the other does not, it's going to be impossible to find a common base. Therefore, the question has no solution.

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Question

Solve .

Answer

First, we gather all the constants on one side of the equation:

Next, we rewrite the equation in exponential form:

Now we can simplify the exponent:

And finally, divide:

$\frac{16}{5}=x$

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