Math AA SL

(a)

The base and the number inside the logarithm are the same, so:

\boldsymbol{{\color{Purple} \log_{4}4 = 1}}

(b)

Using logarithmic laws we get that:

\log_{4}2 + \log_{4}32 = \log_{4}(2 \times 32) = \log_{4}64

Rewriting the expression and using logarithmic laws, we get:

\log_{4}64 = \log_{4}4^{3} = 3\log_{4}4, therefore:

\boldsymbol{{\color{Purple} \log_{4}2 + \log_{4}32 = 3}}

(c)

Using logarithmic laws we get that:

\log_{4}48 - \log_{4}3 = \log_{4}(\frac{48}{3}) = \log_{4}16

Rewriting the expression and using logarithmic laws, we get:

\log_{4}16 = \log_{4}4^{2} = 2\log_{4}4, therefore:

\boldsymbol{{\color{Purple} \log_{4}48 - \log_{4}3 = 2}}

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(a)

Using logarithmic laws we can write that:

\ln40 = \ln(2 \times 20) = \ln2 + \ln20 =

\boldsymbol{{\color{Purple} = x + y}}

(b)

Using logarithmic laws we can write that:

\ln10 = \ln(\frac{20}{2}) = \ln20 - \ln2 =

\boldsymbol{{\color{Purple} = y - x}}

(c)

Using logarithmic laws we can write that:

\ln80 = \ln(4 \times 20) = \ln(2^{2} \times 20) = \ln2^{2} + \ln20) = 2\ln2 + \ln20) =

\boldsymbol{{\color{Purple} = 2x + y}}

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Rearranging and using logarithmic laws, we get:

\ln x^{3} - \ln8 = 12

\ln \left ( \frac{x^{3}}{8} \right ) = 12

Changing the equation from logarithmic to exponential form, we get:

\frac{x^{3}}{8} = e^{12}

Solving for x gives us that:

x = \sqrt[3]{8e^{12}}, hence:

\boldsymbol{{\color{Purple} x = 2e^{4}}}

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(a)

Using logarithmic laws, we get:

-2\ln\left ( \frac{1}{3} \right ) + 3\ln2 = \ln\left ( \frac{1}{3} \right )^{-2} + \ln2^{3} = \ln9 + \ln8, therefore:

\boldsymbol{{\color{Purple} -2\ln\left ( \frac{1}{3} \right ) + 3\ln2 = \ln72}}

(b)

Using our answer from part (a) we can get that:

\ln72= \ln(4x + 8)

72= 4x + 8

\boldsymbol{{\color{Purple} x = 16}}

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Using the nth term of an arithmetic sequence formula we can write that:

u_{5} = u_{1} + (5-1)d

\log_{2}81 = 8 + 4d

Using logarithmic laws and simplifying we get that:

\log_{2}3^{4} = 8 + 4d

4\log_{2}3 = 8 + 4d

d = \log_{2}3 - 2

d = \log_{2}3 - 2\log_{2}2

d = \log_{2}3 - \log_{2}2^{2}

d = \log_{2}3 - \log_{2}4

\boldsymbol{{\color{Purple} d = \log_{2}\left ( \frac{3}{4} \right )}}

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(a)

Using exponential and logarithmic laws we get:

27^{\log_{3}5} = \left ( 3^{3} \right )^{\log_{3}5} =

= 3^{3\log_{3}5} = 3^{\log_{3}5^{3}} = 3^{\log_{3}125} =

\boldsymbol{{\color{Purple} = 125}}

(b)

Using logarithmic laws and rearranging the terms we get:

\log_{6}144 + \log_{6}6 - 2\log_{6}2 = \log_{6}144 - \log_{6}2^{2} + \log_{6}6 =

= \log_{6}144 - \log_{6}4 + 1 = \log_{6}\left ( \frac{144}{4} \right ) + 1 =

=\log_{6}36 + 1 = \log_{6}6^{2} + 1 = 2\log_{6}6 + 1 = 2 + 1 =

\boldsymbol{{\color{Purple} 3}}

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Using logarithmic laws we get:

\log_{2}(2x - 3) + \log_{2}(x + 5) = \log_{2}(3x - 3) + 2

\log_{2}((2x - 3)(x + 5)) = \log_{2}(3x - 3) + 2\log_{2}2

\log_{2}(2x^{2} + 7x - 15) = \log_{2}(3x - 3) + \log_{2}4

\log_{2}(2x^{2} + 7x - 15) = \log_{2}(4(3x - 3))

2x^{2} + 7x - 15 = 4(3x - 3)

Expanding, rearranging for zero and factorizing we get:

2x^{2} - 5x - 3 = 0

(2x + 1)(x - 3) = 0, therefore x = -\frac{1}{2} or x = 3

Checking the original equation we can see that if x = -\frac{1}{2}, then 2x - 3 < 0 and 3x - 3 < 0, hence:

\boldsymbol{{\color{Purple} x = 3}}

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Using exponent laws and rearranging we get that:

\left ( 2^{2} \right )^{x} = 3 - 2^{2} \times 2^{x}

\left ( 2^{x} \right )^{2} + 4 \times 2^{x} - 3 = 0

Using the substitution a = 2^{x}, we get:

a^{2} + 4a - 3 = 0

Using the quadratic formula to solve for a:

a = \frac{-4 \pm \sqrt{4^{2} - 4(1)(-3)}}{2(1)} = \frac{-4 \pm \sqrt{28}}{2} = \frac{-4 \pm 2\sqrt{7}}{2} = -2 \pm \sqrt{7}

Since 2^{x} > 0, we get that:

2^{x} = -2 + \sqrt{7}

Changing from an exponential to a logarithmic form:

\boldsymbol{{\color{Purple} x = \log_{2}(-2 + \sqrt{7})}}

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The information that there is exactly one point on the graph of g where x = 2 means that the equation g(x) = 2 has exactly one solution.

So we will start by writing that:

\log_{(p-2)}(x^{2} - 4x + 13) = 2

Converting from logarithmic to exponential form and working further we get that:

(p - 2)^{2} = x^{2} - 4x + 13

p^{2} - 4p + 4 = x^{2} - 4x + 13

x^{2} - 4x - p^{2} + 4p + 9 = 0

This equation still has exactly one solution, therefore its discriminant is equal to zero.

Hence we get that:

(-4)^{2} - (4)(1)(- p^{2} + 4p + 9) = 0

p^{2} - 4p - 5 = 0

(p + 1)(p - 5) = 0

p = -1 or p = 5

Since p>2, the final answer is:

\boldsymbol{{\color{Purple} p = 5}}

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(a)

d = u_{2} - u_{1} = \ln(2x^{5}) - \ln(x^{5}) = \ln\left ( \frac{2x^{5}}{x^{5}} \right )

\boldsymbol{{\color{Purple} \textbf{Therefore} \, d = \ln2.}}

(b)

Using the nth term of an arithmetic sequence formula we can write that:

u_{6} = u_{1} + (n-1)d

10 = \ln(a^{5}) + (6-1)(\ln2)

10 = \ln(a^{5}) + 5\ln2

10 = \ln(a^{5}) + \ln2^{5}

10 = \ln(2^{5} \times a^{5})

Changing from logarithmic to exponential form:

2^{5} \times a^{5} = e^{10}

a^{5} = \frac{e^{10}}{2^{5}}

\boldsymbol{{\color{Purple} a= \frac{e^{2}}{2}}}

(c)

Let’s begin by finding the common ratio:

r = \frac{u_{2}}{u_{1}} = \frac{\frac{\ln(2x + 7)}{5}}{\ln(2x + 7)}

r = \frac{1}{5}

{\color{Purple} \textbf{Since} \: \boldsymbol{\left | r \right | < 1,} \: \textbf{the sum to infinity exists.}}

(d)

Using the sum of an infinite geometric sequence formula we get that:

S_{\infty} = \frac{\ln(2x + 7)}{1 - \frac{1}{5}} = \frac{\ln(2x + 7)}{\frac{4}{5}} = \frac{5\ln(2x + 7)}{4}

 Using the nth term of an arithmetic sequence formula we get that:

u_{11} = \ln(x^{5}) + (11-1)(\ln2) = \ln(x^{5}) + 10\ln2 = \ln(2^{10} \times x^{5})

The 11th term of the arithmetic sequence is four times the sum to infinity of the geometric sequence, so based on the two expressions that we found above we can write that:

4 \times \frac{5\ln(2x + 7)}{4} = \ln(2^{10} \times x^{5})

Working further we get that:

5\ln(2x + 7) = \ln(2^{10} \times x^{5})

\ln(2x + 7)^{5} = \ln(2^{10} \times x^{5})

(2x + 7)^{5} = 2^{10} \times x^{5}

2x + 7 = 2^{2} \times x

\boldsymbol{{\color{Purple} x = \frac{7}{2}}}

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