Logarithm and Exponential Questions with Answers and Solutions - Grade 12

Step 1: Rewrite the equation so both sides have the same base. We know that any non-zero number to the power of $0$ is $1$:
$$ \left(\frac{1}{2}\right)^{2x + 1} = \left(\frac{1}{2}\right)^0 $$

Step 2: Equate the exponents:
$$ 2x + 1 = 0 $$

Step 3: Solve for $x$:
$$ 2x = -1 \Rightarrow x = -\frac{1}{2} $$

Final Answer: $x = -\frac{1}{2}$

Step 1: Divide all terms in the given equation by $x y$ (assuming $x \neq 0$ and $y \neq 0$):
$$ \frac{x y^m}{x y} = \frac{y x^3}{x y} $$

Step 2: Simplify the exponents using quotient rules:
$$ y^{m - 1} = x^2 $$

Step 3: Take the natural logarithm ($\ln$) of both sides to isolate the exponent containing $m$:
$$ \ln(y^{m - 1}) = \ln(x^2) $$

Step 4: Apply the logarithmic power rule ($ \ln(A^n) = n \ln A $):
$$ (m - 1) \ln y = 2 \ln x $$

Step 5: Solve for $m$ (assuming $\ln y \neq 0$, meaning $y \neq 1$):
$$ m - 1 = \frac{2 \ln x}{\ln y} $$
$$ m = 1 + \frac{2 \ln x}{\ln y} $$

Step 1: Use the log product rule to break down $\log_4(10)$:
$$ \log_4(10) = \log_4(2 \cdot 5) = \log_4(2) + \log_4(5) $$

Step 2: Simplify the first term:
$$ \log_4(2) = \log_4(4^{1/2}) = \frac{1}{2} $$

Step 3: Use the change of base formula on the second term to connect it to the given base 8:
$$ \log_4(5) = \frac{\log_8(5)}{\log_8(4)} $$

Step 4: Substitute the known value $b$ and simplify $\log_8(4)$:
We know $\log_8(4) = \frac{2}{3}$ (because $8^{2/3} = (\sqrt[3]{8})^2 = 4$).
$$ \log_4(5) = \frac{b}{2/3} = \frac{3b}{2} $$

Step 5: Combine the parts:
$$ \log_4(10) = \frac{1}{2} + \frac{3b}{2} = \frac{1 + 3b}{2} $$

Step 1: Simplify the first term recognizing that $216 = 6^3$:
$$ \log_6(216) = \log_6(6^3) = 3 $$

Step 2: Use logarithmic quotient rules to simplify the numerator of the fraction:
$$ \log(42) - \log(6) = \log\left(\frac{42}{6}\right) = \log(7) $$

Step 3: Rewrite the denominator using exponents:
$$ \log(49) = \log(7^2) = 2 \log(7) $$

Step 4: Combine everything into the original expression:
$$ 3 + \frac{\log(7)}{2 \log(7)} $$

Step 5: Cancel $\log(7)$ and calculate the final result:
$$ 3 + \frac{1}{2} = \frac{7}{2} = 3.5 $$

Step 1: Convert the negative exponents into fractions inside the parentheses:
$$ \left( \frac{\frac{1}{3} - \frac{1}{9}}{6} \right)^{1/3} $$

Step 2: Find a common denominator to subtract the fractions in the numerator:
$$ \frac{1}{3} - \frac{1}{9} = \frac{3}{9} - \frac{1}{9} = \frac{2}{9} $$

Step 3: Substitute back and simplify the complex fraction:
$$ \left( \frac{\frac{2}{9}}{6} \right)^{1/3} = \left( \frac{2}{9 \cdot 6} \right)^{1/3} = \left( \frac{2}{54} \right)^{1/3} $$

Step 4: Reduce the fraction and apply the cube root (power of $1/3$):
$$ \left( \frac{1}{27} \right)^{1/3} = \left( \frac{1}{3^3} \right)^{1/3} = \frac{1}{3} $$

Step 1: Use the change of base formula on the second term to write it with base $x$:
$$ \log_a b = \frac{\log_x b}{\log_x a} $$

Step 2: Substitute this back into the original expression:
$$ (\log_x a) \left( \frac{\log_x b}{\log_x a} \right) $$

Step 3: Cancel out the common $\log_x a$ term (assuming $a \neq 1, x \neq 1$):
$$ \log_x b $$

Step 1: Substitute the coordinates $(x=e, y=2)$ into the equation:
$$ 2 = \log_a e $$

Step 2: Rewrite the logarithmic equation in its exponential form:
$$ a^2 = e $$

Step 3: Isolate $a$. Take the square root (since bases of logs must be positive, we only keep the positive root):
$$ a = e^{1/2} = \sqrt{e} $$

Step 1: Use the change of base formula (converting both to natural logs) to rewrite the equation:
$$ \frac{\ln x}{\ln 3} = A \frac{\ln x}{\ln 5} $$

Step 2: Because $x > 0$ and we assume $x \neq 1$ to solve for a general identity, $\ln x \neq 0$. Divide both sides by $\ln x$:
$$ \frac{1}{\ln 3} = \frac{A}{\ln 5} $$

Step 3: Multiply both sides by $\ln 5$ to solve for $A$:
$$ A = \frac{\ln 5}{\ln 3} $$

Step 1: Start from the outermost logarithm (base 10) and rewrite in exponential form:
$$ \log(2 + \log_2(x + 1)) = 10^0 = 1 $$

Step 2: Do it again for the next outermost logarithm (base 10):
$$ 2 + \log_2(x + 1) = 10^1 = 10 $$

Step 3: Isolate the remaining logarithm term:
$$ \log_2(x + 1) = 10 - 2 = 8 $$

Step 4: Rewrite in exponential form (base 2):
$$ x + 1 = 2^8 $$

Step 5: Solve for $x$:
$$ x = 256 - 1 = 255 $$

Check: Plugging 255 back inside yields positive arguments for all logarithms, so the solution is valid.

Step 1: Focus on simplifying the exponential term. Use the power of a power rule backwards: $a^{bc} = (a^b)^c$.
$$ b^{4 \log_b x} = \left(b^{\log_b x}\right)^4 $$

Step 2: Use the inverse property $b^{\log_b x} = x$ to simplify:
$$ \left(b^{\log_b x}\right)^4 = x^4 $$

Step 3: Substitute this back into the original equation:
$$ 2x \cdot x^4 = 486 $$

Step 4: Simplify and solve for $x$:
$$ 2x^5 = 486 $$
$$ x^5 = 243 $$
$$ x = 243^{1/5} = \sqrt[5]{243} = 3 $$

Step 1: Determine the domain.
All arguments of logarithms must be strictly positive:
$x - 1 > 0 \Rightarrow x > 1$
$2x - 1 > 0 \Rightarrow x > 0.5$
$x + 1 > 0 \Rightarrow x > -1$
The intersection of these conditions is $x > 1$. Therefore, the domain is $(1, \infty)$.

Step 2: Use logarithmic power and product rules to condense both sides:
$$ \ln((x - 1)(2x - 1)) = \ln((x + 1)^2) $$

Step 3: Because natural log is one-to-one, we can drop the logs and equate the arguments:
$$ (x - 1)(2x - 1) = (x + 1)^2 $$

Step 4: Expand both sides:
$$ 2x^2 - 3x + 1 = x^2 + 2x + 1 $$

Step 5: Group terms and solve the quadratic equation:
$$ x^2 - 5x = 0 $$
$$ x(x - 5) = 0 $$
Solutions: $x = 0$ and $x = 5$.

Step 6: Verify against the domain.
$x = 0$ is not greater than 1, so it creates negative arguments (extraneous solution). Only $x = 5$ falls in the domain.
Final Answer: $x = 5$

Step 1: Determine the domain.
Inside the square root: $x - 1 \ge 0 \Rightarrow x \ge 1$.
Inside the logarithm: $\sqrt{x - 1} - 2 > 0 \Rightarrow \sqrt{x - 1} > 2 \Rightarrow x - 1 > 4 \Rightarrow x > 5$.
The domain is $x > 5$.

Step 2: The $x$-intercept occurs when $y = 0$:
$$ 0 = 2 \log(\sqrt{x - 1} - 2) $$

Step 3: Divide by 2 and rewrite in exponential form (base 10):
$$ \log(\sqrt{x - 1} - 2) = 0 $$
$$ \sqrt{x - 1} - 2 = 10^0 = 1 $$

Step 4: Isolate the square root and solve for $x$:
$$ \sqrt{x - 1} = 3 $$
$$ x - 1 = 3^2 = 9 $$
$$ x = 10 $$

Step 5: Verify.
$x = 10$ is strictly greater than 5, so it is in the domain.
Final Answer: The $x$-intercept is at $(10, 0)$.

Step 1: Recognize that this is a quadratic equation in disguise. Rewrite $9^x$ in terms of base 3:
$$ 9^x = (3^2)^x = (3^x)^2 $$
So, the equation becomes:
$$ (3^x)^2 - 3^x - 8 = 0 $$

Step 2: Let $y = 3^x$ and rewrite the equation:
$$ y^2 - y - 8 = 0 $$

Step 3: Solve for $y$ using the quadratic formula ($a=1, b=-1, c=-8$):
$$ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-8)}}{2(1)} $$
$$ y = \frac{1 \pm \sqrt{33}}{2} $$

Step 4: Substitute back $y = 3^x$. Because an exponential function $3^x$ must be strictly positive, we must reject the negative root $\left(\frac{1 - \sqrt{33}}{2}\right)$.
$$ 3^x = \frac{1 + \sqrt{33}}{2} $$

Step 5: Take the natural log ($\ln$) of both sides to solve for $x$:
$$ \ln(3^x) = \ln\left( \frac{1 + \sqrt{33}}{2} \right) $$
$$ x \ln(3) = \ln\left( \frac{1 + \sqrt{33}}{2} \right) $$
$$ x = \frac{ \ln\left( \frac{1 + \sqrt{33}}{2} \right) }{ \ln(3) } $$

Step 1: Take the natural log ($\ln$) of both sides:
$$ \ln\left(4^{x - 2}\right) = \ln\left(3^{x + 4}\right) $$

Step 2: Bring down the exponents using the power rule:
$$ (x - 2)\ln(4) = (x + 4)\ln(3) $$

Step 3: Expand both sides:
$$ x \ln(4) - 2 \ln(4) = x \ln(3) + 4 \ln(3) $$

Step 4: Group terms containing $x$ on one side and constants on the other:
$$ x \ln(4) - x \ln(3) = 4 \ln(3) + 2 \ln(4) $$

Step 5: Factor out $x$ and solve:
$$ x (\ln(4) - \ln(3)) = 4 \ln(3) + 2 \ln(4) $$
$$ x = \frac{4 \ln(3) + 2 \ln(4)}{\ln(4) - \ln(3)} $$

Step 6 (Optional): Condense the logarithm for a cleaner final answer:
Numerator: $\ln(3^4) + \ln(4^2) = \ln(81) + \ln(16) = \ln(81 \cdot 16) = \ln(1296)$. Note: $1296 = 6^4$, so this is $4\ln(6)$.
Denominator: $\ln\left(\frac{4}{3}\right)$.
$$ x = \frac{4 \ln(6)}{\ln\left(\frac{4}{3}\right)} $$

Step 1: Rewrite the logarithmic equation in exponential form:
$$ x^{-3/4} = \frac{1}{8} $$

Step 2: To isolate $x$, raise both sides of the equation to the power of the reciprocal, $-\frac{4}{3}$:
$$ \left(x^{-3/4}\right)^{-4/3} = \left(\frac{1}{8}\right)^{-4/3} $$
$$ x^1 = \left(\frac{1}{8}\right)^{-4/3} $$

Step 3: Simplify the right side. The negative exponent flips the fraction:
$$ x = 8^{4/3} $$

Step 4: Evaluate $8^{4/3}$ (cube root first, then 4th power):
$$ x = (\sqrt[3]{8})^4 = 2^4 = 16 $$

Final Answer: $x = 16$