Explore key concepts of logarithmic functions and exponential functions with clear, step-by-step solutions. Learn to solve equations involving logs and exponents, apply the change of base formula, simplify logarithmic expressions, and find intercepts of logarithmic graphs. These examples cover essential skills for algebra, calculus, and real-world science and engineering problems.
Rewrite equation as: \[ \left(\dfrac{1}{2}\right)^{2x + 1} = \left(\dfrac{1}{2}\right)^0 \] Leads to \[ 2x + 1 = 0 \] Solve for \( x \): \[ x = -\dfrac{1}{2} \]
Solve \[ x y^m = y x^3 \] for \( m \).
Divide all terms in the given equation by \( x y \) \[ \dfrac{x y^m}{x y} = \dfrac{y x^3}{x y} \] and simplify: \[ y^{m - 1} = x^2 \]
Take \(\ln\) of both sides: \[ \ln(y^{m - 1}) = \ln (x^2) \] and simplify \[ (m - 1) \ln y = 2 \ln x \]
Solve for \( m \): \[ m = 1 + \dfrac{2 \ln x}{\ln y} \]
Given: \[ \log_8(5) = b \] . Express \( \log_4(10) \) in terms of \( b \).
Use log rule of product to write: \[ \log_4(10) = \log_4(2 \cdot 5) = \log_4(2) + \log_4(5) \] Simplify \[ \log_4(2) = \log_4(4^{1/2}) = \dfrac{1}{2} \] Use change of base formula to write: \[ \log_4(5) = \dfrac{\log_8(5)}{\log_8(4)} = \dfrac{b}{2/3}, \quad \text{since } \log_8(4) = \dfrac{2}{3} \] Simplify to obtain: \[ \log_4(10) = \log_4(2) + \log_4(5) = \dfrac{1 + 3b}{2} \]
Simplify without calculator: \[ \log_{6}(216) + \dfrac{\log(42) - \log(6)}{\log(49)} \]
\[ \log_6(216) + \dfrac{\log(42) - \log(6)}{\log(49)} \] \[ = \log_6(6^3) + \dfrac{\log(42/6)}{\log(7^2)} \] \[ = 3 + \dfrac{\log(7)}{2 \log(7)} = 3 + \dfrac{1}{2} = \dfrac{7}{2} \]
Simplify without calculator: \[ \left( \dfrac{3^{-1} - 9^{-1}}{6} \right)^{1/3} \]
Evaluate the expression:
\[ \left( \dfrac{3^{-1} - 9^{-1}}{6} \right)^{1/3} \] \[ = \left( \dfrac{\dfrac{1}{3} - \dfrac{1}{9}}{6} \right)^{1/3} = \left( \dfrac{\dfrac{6}{27}}{6} \right)^{1/3} \] \[ = \left( {\dfrac{1}{27}} \right)^{1/3} = \left( {\dfrac{1}{3^3}} \right)^{1/3} = \dfrac{1}{3} \]
Express \((\log_{x} a)(\log_{a} b)\) as a single logarithm.
Use change of base formula: \((\log_x a)(\log_a b)\) to write \[ (\log_x a)(\log_a b) = \log_x a \left(\dfrac{\log_x b}{\log_x a}\right) = \log_x b \]
Find \( a \) so that the graph of \( y = \log_{a} x \) passes through the point \( (e, 2) \).
Point \( (e, 2) \) passes through the graph of \( y \), hence: \[ 2 = \log_a e \] Rewrite in exponential form: \[ a^{2} = e \]
Take the \( \ln \) of both sides \[ \ln(a^{2}) = \ln e \] Simplify using the formulas : \( \ln(a^{2}) = 2 \ln a \) and \( \ln e = 1 \): \[ 2 \ln a = 1 \] \[ \ln a = \dfrac{1}{2} \] Rewrite in exponential form: \[ a = e^\left({\dfrac{1}{2}} \right) = \sqrt{e} \]
Find the constant \( A \) such that \[ \log_{3} x = A \log_{5} x \] for all \( x > 0 \).
Use the change of base formula with natural logarithms to rewrite the given equation as follows: \[ \dfrac{\ln(x)}{\ln(3)} = A \dfrac{\ln(x)}{\ln(5)} \]
Solve for \( A \): \[ A = \dfrac{\ln(5)}{\ln(3)} \]
Solve for \( x \) the equation: \[ \log \left( \log \left( 2 + \log_2 (x + 1) \right) \right) = 0 \]
Rewrite given equation as: \[ \log \bigl( \log (2 + \log_{2}(x + 1)) \bigr) = \log(1), \] since \(\log(1) = 0\). \[ \log (2 + \log_{2}(x + 1)) = 1 \]
Rewrite in exponential form: \[ 2 + \log_{2}(x + 1) = 10 \]
Write term with the \( \log \) on one side: \[ \log_{2}(x + 1) = 8 \]
Write in exponential form: \[ x + 1 = 2^{8} \]
Solve for \( x \): \[ x = 2^{8} - 1 = 255 \]
Solve for \( x \) the equation \[ 2 x \cdot b^{4 \log_b x} = 486 \]
Note that: \[ b^{4 \log_b x} = (b^{\log_b x})^4 = x^4 \]
Hence, The given equation may be written as: \[ 2x \cdot x^4 = 486 \]
which simplifies to \[ x^5 = 243 \] and then \[ x = 243^{\dfrac{1}{5}} = 3 \]
Solve for \( x \) the equation \[ \ln(x - 1) + \ln(2x - 1) = 2 \ln(x + 1) \]
Use the product rule to write the left side as: \[ \ln(x - 1) + \ln(2x - 1) = \ln ((x - 1)(2x - 1)) \]
Rewrite the give equation as: \[ \ln ((x - 1)(2x - 1)) = \ln((x + 1)^2) \] which gives the algebraic equation \[ (x - 1)(2x - 1) = (x + 1)^2 \]
Expand, simplify and rewrite the equation in standard form: \[ x^2 - 5 x = 0 \] Solve for \( x \) to obtain two solutions: \[ x = 0 \quad \text{and} \quad x = 5 \]
Check solutions: \(x = 0 \) is NOT in the domain of the left side of the equation and hence is not a solution to the given equation.
\( x = 5 \) is a solution of the given equation. (Check it as an exercie)
Find the \( x \)-intercept of the graph of \[ y = 2 \log\left(\sqrt{x - 1} - 2\right) \]
The \( x \)-intercept is found by solving the equation : \[ 0 = 2 \log(\sqrt{x - 1} - 2) \]
Divide both sides by 2: \[ \log(\sqrt{x - 1} - 2) = 0 \]
Rewrite in exponential form: \[ \sqrt{x - 1} - 2 = 10^0 = 1 \]
Rewrite the above equation as: \[ \sqrt{x - 1} = 3 \]
Raise both sides to the power 2: \[ (x - 1) = 3^2 \] Solve for \( x \) \[ x = 10 \] Note: check that \( x = 10 \) is in the domain of the function.
Solve for \( x \) the equation \[ 9^x - 3^x - 8 = 0 \]
Given: \[ 9^x - 3^x - 8 = 0 \] Note that: \[ 9^x = (3^x)^2 \]
Equation may be written as \[ (3^x)^2 - 3^x - 8 = 0 \]
Let \( y = 3^x \) and rewrite equation in terms of \( y\) as : \[ y^2 - y - 8 = 0 \]
Solve for \( y \) to obtain two solutions: \[ \quad y = \dfrac{1 + \sqrt{33}}{2} \quad \text{and} \quad y = \dfrac{1 - \sqrt{33}}{2} \]
Since \( y = 3^x \) (positive), the only acceptable solution is : \[ y = \dfrac{1 + \sqrt{33}}{2} \]
Substitute \( y \) by \( 3^x \) to obtain the equation: \[ 3^x = \dfrac{1 + \sqrt{33}}{2} \]
Take the \( \ln \) of both sides: \[ \ln(3^x) = \ln\left( \dfrac{1 + \sqrt{33}}{2} \right) \]
Simplify and solve: \[ x = \dfrac{ \ln\left( \dfrac{1 + \sqrt{33}}{2} \right) }{ \ln(3) } \]
Solve for \( x \) the equation: \[ 4^{x - 2} = 3^{x + 4} \] .
Given: \[ \quad 4^{x - 2} = 3^{x + 4} \] Take \(\ln\) of both sides: \[ \ln\left(4^{x - 2}\right) = \ln\left(3^{x + 4}\right) \] Simplify: \[ (x - 2)\ln 4 = (x + 4)\ln 3 \] Expand: \[ x \ln 4 - 2 \ln 4 = x \ln 3 + 4 \ln 3 \] Group like terms: \[ x \ln 4 - x \ln 3 = 4 \ln 3 + 2 \ln 4 \] Solve for \(x\): \[ x = \dfrac{4 \ln 3 + 2 \ln 4}{\ln 4 - \ln 3} = \dfrac{\ln\left(3^4 \cdot 4^2\right)}{\ln\left(\dfrac{4}{3}\right)} \] \[ = \dfrac{\ln\left(3^4 \cdot 2^4\right)}{\ln\left(\dfrac{4}{3}\right)} = \dfrac{4 \ln 6}{\ln\left(\dfrac{4}{3}\right)} \]
If \(\log_{x}\left(\dfrac{1}{8}\right) = -\dfrac{3}{4}\), then what is \(x\)?
Rewrite the given equation using exponential form: \[ x^{-\dfrac{3}{4}} = \dfrac{1}{8} \]
Raise both sides of the above equation to the power \(-\dfrac{4}{3}\): \[\left(x^{-\dfrac{3}{4}}\right)^{-\dfrac{4}{3}} = \left(\dfrac{1}{8}\right)^{-\dfrac{4}{3}}\]
simplify and solve for \( x \): \[ x = 8^{\left(\dfrac{4}{3}\right)} = 2^{4} = 16 \]