\[ 32^{x-1} = 4^{4x}\]
\[ 2^{5(x-1)} = 2^{8x}\]
\[ 2^{5x-5} = 2^{8x}\]
\[ 5x - 5 = 8x\]
\[ x = -\frac{5}{3}\]
Closea) 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.
a) 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) 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: \( x + 2y \)
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) 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, \( x = \frac{1}{3} \)
CloseWe start by simplifying the first equation:
\[ 5^x \cdot 5^{\frac{11}{4}y} = 5^{x + \frac{11}{4}y} \]
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:
\[ y = 4 \cdot \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \]
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 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:
\[8^{3b} = 3^{3}\]
\[3b \ln 8 = 3 \ln 3\]
\[b \ln 8 = \ln 3\]
\[b = \frac{\ln 3}{\ln 8}\]
CloseTo solve the equation \( 4^{x^2 - 2} = \left( \frac{1}{16} \right)^{3x + 1} \):
Notice that \( 4 \) and \( 16 \) are both powers of \( 2 \):
\[ 4 = 2^2 \quad \text{and} \quad 16 = 2^4 \]
Therefore, \( \frac{1}{16} = 2^{-4} \).
\[ 4^{x^2 - 2} = \left( \frac{1}{16} \right)^{3x + 1} \implies (2^2)^{x^2 - 2} = (2^{-4})^{3x + 1} \]
\[ 2^{2(x^2 - 2)} = 2^{-4(3x + 1)} \]
This simplifies to:
\[ 2^{2x^2 - 4} = 2^{-12x - 4} \]
Since the bases are the same, the exponents must be equal:
\[ 2x^2 - 4 = -12x - 4 \]
\[ 2x^2 - 4 + 12x + 4 = 0 \implies 2x^2 + 12x = 0 \]
Factor out \( 2x \):
\[ 2x(x + 6) = 0 \]
This gives two solutions:
\[ 2x = 0 \implies x = 0, \]
and
\[ x + 6 = 0 \implies x = -6 \]
The solutions to the equation are \( x = 0 \) and \( x = -6 \).
Closea) Using the power rule of logarithms, \( \log_b a^n = n \log_b a \):
\[ \log_3 x^4 = 4 \log_3 x = 4z \]
b) Using the reciprocal rule of logarithms, \( \log_b \left( \frac{1}{a} \right) = -\log_b a \):
\[ \log_3 \left( \frac{1}{x} \right) = -\log_3 x = -z \]
c) First, express \( \log_{9} x \) in terms of base 3. Since \( 9 = 3^2 \), we can use the change of base formula:
\[ \log_{9} x = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2} = \frac{z}{2} \]
Closea) To solve the equation \( \log_3 x = \log_9 (x + 8) \), we can do the following:
Notice that \( 9 \) is a power of \( 3 \):
\[ 9 = 3^2 \]
Therefore, we can express \( \log_9 (x + 8) \) in terms of base \( 3 \) using the change of base formula:
\[ \log_9 (x + 8) = \frac{\log_3 (x + 8)}{\log_3 9} = \frac{\log_3 (x + 8)}{2} \]
Substitute the expression for \( \log_9 (x + 8) \) into the original equation:
\[ \log_3 x = \frac{\log_3 (x + 8)}{2} \]
\[ 2 \log_3 x = \log_3 (x + 8) \]
The left side can be rewritten using the power rule \( \log_b a^n = n \log_b a \):
\[ \log_3 x^2 = \log_3 (x + 8) \]
Since the logarithms are equal and have the same base, their arguments must be equal:
\[ x^2 = x + 8 \]
Rearrange the equation to standard quadratic form:
\[ x^2 - x - 8 = 0 \]
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -8 \):
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-8)}}{2(1)} = \frac{1 \pm \sqrt{1 + 32}}{2} = \frac{1 \pm \sqrt{33}}{2} \]
Since the logarithm is only defined for positive arguments, we must ensure that \( x > 0 \) and \( x + 8 > 0 \). Both solutions \( \frac{1 + \sqrt{33}}{2} \) and \( \frac{1 - \sqrt{33}}{2} \) must be checked:
Therefore, the only valid solution is:
\[ x = \frac{1 + \sqrt{33}}{2} \]
The solution to the equation \( \log_3 x = \log_9 (x + 8) \) is:
\[ x = \frac{1 + \sqrt{33}}{2} \]
Close