Exponents and Logarithms

SAT Subject Test in Math II · Learn by Concept

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SAT Subject Test in Math II › Exponents and Logarithms

1 - 10

1

Solve for :

$5^{x} = 432$

Give the solution to the nearest hundredth.

$x \approx 3.77$

CORRECT

$x \approx 1.94$

0

$x\approx 3.37$

0

$x \approx 2.44$

0

$x \approx 2.86$

0

Explanation

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

$5^{x} = 432$

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

$x \log 5= \log 432$

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

2

Solve for :

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

Give your answer to the nearest hundredth.

$k \approx 3.51$ '

CORRECT

$k \approx 1.43$

0

$k \approx 3.61$

0

$k \approx 1.69$

0

$k \approx 1.89$

0

Explanation

Take the common logarithm of both sides and solve for :

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

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

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

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

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

3

Solve for :

$10^{t} + 1 = e$

Give your answer to the nearest hundredth.

$t \approx 0.24$

CORRECT

The equation has no solution.

0

$t \approx 0.46$

0

$t \approx 0.42$

0

$t \approx -0.57$

0

Explanation

$10^{t} + 1 = e$

$10^{t} = e - 1$

Take the common logarithm of both sides and solve for :

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

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

4

Solve for :

$e^{x} = \pi$

Give your answer to the nearest hundredth.

$x \approx 1.14$

CORRECT

$x \approx 0.87$

0

$x \approx 0.50$

0

$x \approx 2.01$

0

The equation has no solution.

0

Explanation

Take the natural logarithm of both sides and solve for :

$e^{x} = \pi$

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

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

5

To the nearest hundredth, solve for :

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

$x \approx 1.45$

CORRECT

The equation has no solution.

0

$x\approx 7.17$

0

$x \approx 3.85$

0

$x \approx 2.31$

0

Explanation

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

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

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

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

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

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

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

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

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

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

6

To the nearest hundredth, solve for :

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

$x \approx 3.03$

CORRECT

$x \approx 1.20$

0

$x \approx -1.68$

0

$x \approx 6.32$

0

$x \approx 4.48$

0

Explanation

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

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

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

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

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

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

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

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

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

7

Solve for :

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

CORRECT

0

0

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

0

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

0

Explanation

$9 = 3^{2}$ 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,

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

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

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

8

Solve for :

$100 ^{x}= 66$

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

CORRECT

$x = \log 33$

0

$x = \log 64$

0

$x = 2- \log 66$

0

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

0

Explanation

Take the common logarithm of both sides:

$100 ^{x}= 66$

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

$x \log 100 = \log 66$

$x \cdot 2 = \log 66$

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

9

Solve for :

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

CORRECT

0

0

0

0

Explanation

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,

10

Solve for :

$81^{x-3}=27^{2x}$

CORRECT

0

0

0

Explanation

In order to solve this problem, rewrite both sides of the equation in terms of raising to an exponent.

Since, , we can write the following:

$81^{x-3}=(3^4)^{x-3}=3^{4x-12}$

Since , we can write the following:

$27^{2x}=(3^3)^{2x}=3^{6x}$

Now, we can solve for with the following equation:

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