Let \(a\), \(b\), \(x\), and \(y\) be real numbers and let \(n\) be an integer with \(n \geq 2\). The rules for exponents are
\(a^{\frac{1}{n}}=\sqrt[n]{a}\)
\(a^0=1\), if \(a\neq0\)
\(a^{-x}=\dfrac{1}{a^x}\), if \(a\neq0\)
\(a^xa^y=a^{x+y}\)
\(\dfrac{a^x}{a^y}=a^{x-y}\), if \(a\neq0\)
\((a^x)^y=a^{xy}\)
\(a^x\cdot b^x=(ab)^x\)
\(\dfrac{ a^x}{b^x}=\left(\dfrac{a}{b}\right)^x\), if \(b\neq0\)
Also, \(0^0\) is not defined, if it is encountered using any of the above formulae.
Let \(x\) and \(y\) be positive real numbers. Let \(a\) be a positive real number with \(a \neq 1\). The equation \(y= a^x\) is equivalent to \(\log_a y = x\). The rules for logarithms are
\(\log_a1=0\)
\(\log_a a = 1\)
\(\log_a(xy)=\log_ax+\log_ay\)
\(\log_a\left(\dfrac{x}{y}\right)=\log_ax-\log_ay\)
\(\log_a(x^y)=y\log_ax\)
\(\log_a(a^x)=x\)
\(a^{\log_ax}=x\)
\(\log_ax=\dfrac {1}{\log_xa}\), where \(x \neq 1\)
\(\log_yx=\dfrac{\log_ax}{\log_ay}\)
Given \(2\log_5(x-3y)=\log_5(2x)+\log_5(2y)\), calculate the ratio \(\dfrac{x}{y}\).
Solution
First, we note that in the original equation, if the three logarithmic terms are to be defined, then their arguments must be positive. So \(x>0\), \(y>0\), and \(x>3y\). Now \[\begin{align*} 2\log_5(x-3y)&=\log_5(2x)+\log_5(2y)\\ \log_5(x-3y)^2&=\log_5(4xy).\end{align*}\]
Now since the log function takes on each value in its range only once, we have that \[\begin{align*} (x-3y)^2&=4xy\\ x^2-6xy+9y^2 &= 4xy\\ x^2-10xy+9y^2 &= 0\\ (x-y)(x-9y) &= 0\end{align*}\] So \(\dfrac{x}{y}=1\) or \(\dfrac{x}{y} = 9\). But from our restrictions we know that \(\dfrac{x}{y}>3\), and so \(\dfrac{x}{y}=9\).
Given that \(m\) and \(k\) are integers, find all values of \(m\) and \(k\) satisfying the equation \[9(7^k+7^{k+2})=5^{m+3}+5^m\]
Solution
We factor both sides of this equation to arrive at \[\begin{align*} 9(1+7^2)7^k&=5^m(1+5^3)\\ 9(50)7^k &= 5^m(126) \\ 3^2\cdot 2\cdot 5^2\cdot 7^k&=5^m\cdot 2\cdot 3^2\cdot 7\end{align*}\] Now since both sides of this equation are products of primes, and integers have unique prime factorizations, it follows that \(m=2\) and \(k=1\) is the only solution.
Determine the points of intersection of the curves \(y=\log_{10}(x-2)\) and \(y=1-\log_{10}(x+1)\).
Solution
Again the arguments of the logarithmic functions, \(x-2\) and \(x+1\), must be positive, which implies that \(x>2\). Now \[\begin{align*} \log_{10}(x-2) &=1-\log_{10}(x+1)\\ \log_{10}(x-2)+\log_{10}(x+1)&=1\\ \log_{10}[(x-2)(x+1)]&=1\\ (x-2)(x+1)&=10\\ x^2-x-2&=10\\ x^{2}-x-12&=0\\ (x-4)(x+3)&=0\end{align*}\] So \(x=4\) or \(x=-3\), but from our restriction \(x>2\) and so \(x = 4\). The point of intersection is \((4,\log_{10}2)\) or \((4,1-\log_{10}5)\). Since \(\log_{10}2+\log_{10}5=1\), these are equivalent answers.
Determine all values of \(x\) such that \(\log_2(9-2^x)=3-x\).
Solution
Once again the argument of the logarithm must be positive, implying that \(9>2^x\). \[\begin{align*} \log_2(9-2^x)&=3-x\\ 9-2^x=2^{3-x}&=\dfrac{8}{2^x}\end{align*}\]
Substituting \(y=2^x\) we have \[\begin{align*} 9-y&=\dfrac{8}{y}\\ y^2-9y+8&=0 \\ (y-1)(y-8)&=0\end{align*}\] Thus, \(y=1\) or \(y=8\). Since \(y=2^x\), we obtain the corresponding values \(x=0\) or \(x=3\). Both of these values satisfy the restriction \(9 >2^x\) and so both are valid solutions.
The graph of \(y=m^x\) passes through the points \((2,5)\) and \((5,n)\). What is the value of \(mn\)?
Solution
From the given information we have that \(m^2=5\) and \(n=m^5\). Thus, \(m=\pm\sqrt{5}\) with corresponding values \(n=(\pm\sqrt {5})^5\). Therefore, \(mn=(\sqrt{5})^6=125\).
Determine the values of \(x\) such that \(\log_x2+\log_x4+\log_x8=1\).
Determine the values of \(x\) such that \(12^{2x+1}=2^{3x+7}\cdot3^{3x-4}\).
What is the sum of the following series?
\[\log_{10}\dfrac{3}{2}+\log_{10}\dfrac{4}{3}+\log_{10}\dfrac{5}{4}+\cdots+\log_{10}\dfrac{200}{199}\]
Given that \(x^3y^5=2^{11}\cdot3^{13}\) and \(\dfrac{x}{y^2}=\dfrac{1}{27}\), solve for \(x\) and \(y\).
Given that \(\log_83=k\), express \(\log_818\) in the form \(ak+b\) where \(a\) and \(b\) are rational numbers.
Determine the point(s) of intersection of the graphs of \(y=\log_2(2x)\) and \(y=\log_4x\).
The points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) lie on the graph of \(y=\log_ax\). A horizontal line is drawn through the midpoint of \(AB\) such that it intersects the curve at the point \(C(x_3,y_3)\). Show that \((x_3)^2=x_1x_2\).
The graph of the function \(y=ax^r\) passes through the points \((2,1)\) and \((32,4)\). Calculate the value of \(r\).
Given that \(2^{x+3}+2^x=3^{y+2}-3^y\) and \(x\) and \(y\) are integers, determine the values of \(x\) and \(y\).
Given that \(f(x)=2^{4x-2}\), calculate, in simplest form, \(f(x)\cdot f(1-x)\).
Find all values of \(x\) such that \(\log_5(x-2)+\log_5(x-6)=2\).
Let \(x\) be a positive real number with \(x \neq 1\). Prove that \(a\), \(b\), and \(c\) are three numbers that form a geometric sequence if and only if \(\log_xa\), \(\log_xb\) and \(\log_xc\) form an arithmetic sequence.
Determine all real values of \(x\) for which \[3^{x+2} +2^{x+2} +2^x = 2^{x+5} +3^x\]
Determine all real numbers \(x\) for which \[(\log_{10} x)^{\log_{10}(\log_{10} x)} = 10000\]
Determine all real numbers \(x > 0\) for which \[\log_4 x-\log_x 16 = \dfrac{7}{6} -\log_x 8\]