a) This is just a simple \(2^4\).
b) Taking two to the power of a fraction will result in a square root. Thus, we need to first write it as \( \sqrt{2^3} \), and then as \((2^3)^{1/2}\), resulting in the final answer of \(2^{3/2}\).
c) When a number is taken to a negative power then a fraction is created, thus we need to find to which power 2 needs to be raised to result in 128. The answer for that is 7, so \(2^7\) is equal to 128. Therefore, \(-2^{-7}\) is the answer.
\[ 32^{x-1} = 4^{4x}\]
\[ 2^{5(x-1)} = 2^{8x}\]
\[ 2^{5x-5} = 2^{8x}\]
\[ 5x - 5 = 8x\]
\[ x = -\frac{5}{3}\]
Closea) Using the rule \( \ln(x/y) = \ln(x) - \ln(y) \), we get:
\[ \ln(4) = \ln(20) - \ln(5) \]
So, the final answer is: \( x - y \)
b) Using the same rule \( \ln(x/y) = \ln(x) - \ln(y) \), we get:
\[ \ln\left(\frac{1}{4}\right) = \ln(5) - \ln(20) \]
Thus, the final answer is: \( y - x \)
c) Using the rule \( y \ln(x) = \ln(x^y) \), we have:
\[ 2 \ln(5) = \ln(25) \]
So, the final answer is: \( x + 2y \)
\[ \log_5 2 + \log_5 10 = \log_5 (2 \cdot 10) = \log_5 20 \]
Thus, the equation simplifies to:
\[ \log_5 y + 1 = \log_5 20 \]
Rewrite \( \log_5 y + 1 \) as \( \log_5 (5y) \), so:
\[ \log_5 (5y) = \log_5 20 \]
Equate the arguments:
\[ 5y = 20 \]
Therefore:
\[ y = \frac{20}{5} = 4 \]
a) To express \( 3 \ln 3 - \ln 9 \) in the form \( \ln k \):
We know that \( 3\ln 3 = \ln (3^3) = \ln 27 \), so:
\[ 3 \ln 3 - \ln 9 = \ln 27 - \ln 9 = \ln \left( \frac{27}{9} \right) = \ln 3 \].
Hence, \( k = 3 \).
b) Given the equation \( 3 \ln 3 - \ln 9 = -\ln x \):
From part (a), we found that \( 3 \ln 3 - \ln 9 = \ln 3 \), so:
\[ \ln 3 = -\ln x \]
\[ -\ln 3 = \ln x \]
\[ \ln 3^{-1} = \ln x \]
\[ \ln \frac{1}{3} = \ln x \]
Thus, \( x = \frac{1}{3} \)
Closea) Both numerator and denominator contain only multiplication and both unknowns. Thus, operations on power can be performed for each unknown, such as:
\[ \frac{32a^2 b^3}{8\sqrt{a} b^5} = \frac{32a^2 b^3}{8a^{1/2} b^5} = \frac{4a^{3/2}}{b^2} \]
b) Knowing that \( b = 2a \) we can plug it into the final fraction obtained in part a.
\[ \frac{4a^{3/2}}{b^2} = \frac{4a^{3/2}}{(2a)^2} = a^{-1/2} \]
\[ a^{-1/2} = 8 \]
\[ \frac{1}{\sqrt{a}} = 8 \Rightarrow 1 = 8\sqrt{a} \Rightarrow a = \frac{1}{64} \]
\[ b = \frac{1}{32} \]
Closea) (i) To find the x-intercept, set \( f(x) = 0 \):
\[\frac{2 \ln(x+6)}{3} = 0 \]
Solving for \( x \):
\[ \ln(x+6) = 0 \implies x+6 = e^0 \implies x+6 = 1 \implies x = -5 \]
Thus, the x-intercept is \( x = -5 \).
a) (ii) To find the y-intercept, set \( x = 0 \):
\[ f(0) = \frac{2 \ln(0+6)}{3} = \frac{2 \ln 6}{3} \]
Thus, the y-intercept is \( y = \frac{2 \ln 6}{3} \).
b) The vertical asymptote occurs where the argument of the logarithm is zero or negative:
\[ x+6 \leq 0 \implies x \leq -6 \]
Thus, the vertical asymptote is at \( x = -6 \).
c) The function is defined where the argument of the logarithm is positive:
\[ x+6 > 0 \implies x > -6 \]
Thus, the domain of \( f(x) \) is \( x > -6 \), or in interval notation: \( (-6, \infty) \).
a) The answer here is simply 3, as \( \log_4 64 = 3 \) since \( 4^3 = 64 \).
b) We start by simplifying the expression inside the logarithm:
\[ \log_4 \left( \frac{64^{3x-1}}{16^{2y+2}} \right) = \log_4 \left( \frac{4^{3(3x-1)}}{4^{2(2y+2)}} \right) = \log_4 \left( \frac{4^{9x-3}}{4^{4y+4}} \right) = \log_4 \left( 4^{9x-4y-7} \right) = 9x - 4y - 7 \]
Thus, \( a = 9 \), \( b = -4 \), and \( c = -7 \).
a) Given \( u_5 = \ln 3 \) and common difference \( d = \ln 3 \). To find \( u_7 \), use the formula for the nth term of an arithmetic sequence:
\[ u_n = u_1 + (n - 1) \times d \]
We have:
\[ u_5 = u_1 + 4 \times d \]
Given \( u_5 = \ln 3 \) and \( d = \ln 3 \), so:
\[ \ln 3 = u_1 + 4 \times \ln 3 \]
\[ u_1 = \ln 3 - 4 \times \ln 3 = -3 \times \ln 3\]
Now find \( u_7 \):
\[ u_7 = u_1 + 6 \times d \]
\[ u_7 = -3 \times \ln 3 + 6 \times \ln 3 = 3 \times \ln 3 = \ln 81 \]
Thus, \( u_7 = \ln 81 \).
b) We already found \( u_1 \) in part (a):
\[ u_1 = -3 \times \ln 3 \]
c) The sum \( S_n \) of the first \( n \) terms of an arithmetic sequence is given by:
\[ S_n = \frac{n}{2} \times (2u_1 + (n - 1) \times d) \]
Substitute \( u_1 = -3 \times \ln 3 \) and \( d = \ln 3 \):
\[ S_n = \frac{n}{2} \times \left(2 \times (-3 \times \ln 3) + (n - 1) \times \ln 3 \right) \]
\[ S_n = \frac{n}{2} \times \left(-6 \times \ln 3 + (n - 1) \times \ln 3 \right) \]
\[ S_n = \frac{n}{2} \times \left((n - 7) \times \ln 3 \right) \]
\[ S_n = \ln 3 \times \frac{n \times (n - 7)}{2} \]
\[ S_n = \ln \sqrt{3} (n^2 - 7n) \]
d) To find the x-coordinates of the intersection points of \( g(x) = \ln \sqrt{3} \times (x^2 - 7x) \) and \( h(x) = 2x \times \ln 3 \), set \( g(x) = h(x) \):
\[ \ln \sqrt{3} \times (x^2 - 7x) = 2x \times \ln 3\]
Divide both sides by \( \ln 3 \):
\[ \frac{\ln \sqrt{3}}{\ln 3} \times (x^2 - 7x) = 2x \]
Since \( \ln \sqrt{3} = \frac{1}{2} \times \ln 3 \), we have:
\[ \frac{\frac{1}{2} \times \ln 3}{\ln 3} \times (x^2 - 7x) = 2x \]
\[ \frac{1}{2} \times (x^2 - 7x) = 2x \]
\[ x^2 - 7x = 4x \]
\[ x^2 - 11x = 0 \]
\[ x(x - 11) = 0 \]
Thus, the x-coordinates of the points of intersection are \( x = 0 \) and \( x = 11 \).
We start by simplifying the first equation:
\[ 5^x \times 5^{\frac{11}{4}y} = 5^{x + \frac{11}{4}y} \]
Since \(\sqrt{125} = 125^{1/2} = (5^3)^{1/2} = 5^{3/2}\), we can write the equation as:
\[ 5^{x + \frac{11}{4}y} = 5^{3/2} \]
By equating the exponents of the same base, we get:
\[ x + \frac{11}{4}y = \frac{3}{2} \quad \text{(Equation 1)} \]
Next, we simplify the second equation. Recall that \( 25 = 5^2 \), so we can rewrite the equation as:
\[ (5^2)^x = 5^{\frac{y}{2}} \Rightarrow 5^{2x} = 5^{\frac{y}{2}} \]
By equating the exponents of the same base, we get:
\[ 2x = \frac{y}{2} \Rightarrow y = 4x \quad \text{(Equation 2)} \]
Now, substitute \( y = 4x \) into Equation 1:
\[ x + \frac{11}{4}(4x) = \frac{3}{2} \]
\[ x + 11x = \frac{3}{2} \]
\[ 12x = \frac{3}{2} \]
\[ x = \frac{3}{2 \cdot 12} = \frac{3}{24} = \frac{1}{8} \]
Using \( y = 4x \), we get:
\[ y = 4 \times \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \]
CloseTo find the value of \( a \), solve the equation \( \log_a(\sqrt{32}x - 2x^2) = 2 \) which implies:
\[\sqrt{32}x - 2x^2 = a^2\]
For this quadratic equation to have exactly one solution, the discriminant must be zero. The quadratic equation is:
\[ -2x^2 + \sqrt{32}x - a^2 = 0 \]
Discriminant \( \Delta \) of this quadratic equation is given by:
\[ \Delta = ( \sqrt{32})^2 - 4 \times (-2) \times (-a^2) = 32 - 8 a^2 \]
Set \( \Delta = 0 \) for the equation to have one solution:
\[ 32 - 8 a^2 = 0 \implies a^2 = 4 \implies a = \pm 2 \]
However, we know that \( a \) is bigger than 0, so the final value is \( a = 2 \).
Let's simplify the equation:
\[24^{3b} = 27^{b+1}\]
\[3^{3b} * 3^{3b} = 27^{b+1}\]
\[3^{3b} * 8^{3b} = 3^{3^{b+1}}\]
\[3^{3b} * 8^{3b} = 3^{3b + 3}\]
\[3^{3b} * 8^{3b} = 3^{3b} * 3^3\]
Now, \(3^{3b}\) will cancel out:
\[8^{3b} = 3^{3}\]
\[3b \ln 8 = 3 \ln 3\]
\[b \ln 8 = \ln 3\]
\[b = \frac{\ln 3}{\ln 8}\]
Closea) The common ratio \( r \) is:
\[r = \frac{\ln x^2}{2 \ln x^2} = \frac{2 \ln x}{4 \ln x} = \frac{1}{2}\]
b) Using the common ratio \( r = \frac{1}{2} \) and the formula for the general term, we get:
\[u_n = u_1 * r^{n-1}\]
\[u_n = 2 \ln x^2 * (\frac{1}{2})^{n-1}\]
\[u_n = 4 \ln x * (\frac{1}{2})^{n-1}\]
\[u_n = (\frac{1}{2})^{-2} \ln x * (\frac{1}{2})^{n-1}\]
\[u_n = (\frac{1}{2})^{n-3} \ln x\]
\[u_n = 2^{3-n} \ln x\]
c) From part (b) we learned that:
\[u_n = 2^{3-n} \ln x\]
So, this is going to be just a sum to infinity of the series in this question, so we can apply the formula for the sum to infinity:
\[\frac{u_1}{1-r} = 24\]
\[\frac{2 \ln x^2}{1-\frac{1}{2}} = 24\]
\[\frac{4 \ln x}{\frac{1}{2}} = 24\]
\[8 \ln x = 24\]
\[x = e^3\]
CloseLet's use the substitution \( y = 4^x \). Then:
\[ 4^{2x} = (4^x)^2 = y^2 \]
\[ \frac{1}{4} \times 4^{x+2} = \frac{1}{4} \times 16y = 4y \]
The original equation becomes:
\[ y^2 + 4y = 2 \]
Rearrange into standard quadratic form:
\[ y^2 + 4y - 2 = 0 \]
Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 4 \), and \( c = -2 \):
\[ y = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-2)}}{2 \times 1} \]
\[ y = \frac{-4 \pm \sqrt{16 + 8}}{2} = \frac{-4 \pm \sqrt{24}}{2} \]
Since \( \sqrt{24} = 2\sqrt{6} \), we get:
\[ y = \frac{-4 \pm 2\sqrt{6}}{2} = -2 \pm \sqrt{6} \]
We discard the negative solution as \( y = 4^x \) must be positive. Thus:
\[ y = -2 + \sqrt{6} \]
So, \( 4^x = -2 + \sqrt{6} \). Taking logarithms to find \( x \):
\[ x = \log_4 \left(-2 + \sqrt{6}\right) \]
Thus, we have \( a = 4 \) and \( b = 6 \).