We have \(\log_x(2\cdot4\cdot8)=1\) which implies that \(\log_x(64)=1\). Therefore, \(x=64\).
Since \(12=2^2\cdot3\) it follows that \[\begin{align*} 12^{2x+1}&=2^{3x+7}\cdot3^{3x-4} \\ 2^{2(2x+1)}\cdot 3^{{2x+1}}&=2^{{3x+7}}\cdot 3^{{3x-4}}\\ 2^{2(2x+1)-3x-7} &=3^{3x-4-2x-1}\\ 2^{x-5} &=3^{x-5}\end{align*}\] The graphs of \(y=2^{x-5}\) and \(y=3^{x-5}\) intersect only at \(x=5\) and \(y=1\). (Since \(2^z = 3^z\) only for \(z=0\).) Therefore, \(x=5\) is the only solution.
This expression equals \(\log_{10}\bigg(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot\cdots\cdot\dfrac{200}{199}\bigg)=\log_{10}\displaystyle\left(\dfrac{200}{2} \right)= \log_{{10}}100 = 2\).
For the first equation we know that \(x, y \neq 0\). The second equation states \(xy^{-2}=3^{-3}\) which gives us \(x^3y^{-6}=3^{-9}\) after cubing both sides. Since neither side of this new equation is \(0\) we can divide the first equation by this new equation to eliminate \(x\). \[\begin{align*} \dfrac{x^3y^5}{x^3y^{-6}}&=\dfrac{2^{11}3^{13}}{3^{-9}}\\ y^{11}&=2^{11}\cdot3^{22}\\ y^{11} & = 2^{11} \cdot (3^2)^{11} \\ y^{11} & = (2 \cdot 9)^{11} \\ y&=18.\end{align*}\] Therefore, \(x=\dfrac{y^{2}}{27}=12\).
\(\log_8(18)=\log_82+\log_89 = \dfrac{1}{3} + \log_8 3^2 = \dfrac{1}{3} + 2 \log_8 3=\dfrac{1}{3}+2k=2k+\dfrac{1}{3}\)
Solution 1
We express the logarithms in exponential form to arrive at: \(2x=2^y\) and \(x=4^y\). Thus, \[\begin{align*} 2^y &=2(4^y)\\ 2^y &=2(2^{2y})\\ 2^y&=2^{2y+1}\\ y&=2y+1\\ y&=-1\end{align*}\] Thus, \(x=4^{-1} = \dfrac{1}{4}\). Therefore, the point of intersection is \(\left( \dfrac{1}{4}, -1\right)\).
Solution 2
Substituting one equation into the other we obtain \[\begin{align*} \log_2 2x &= \log_4x \\ \dfrac{\log 2x}{\log 2} &= \dfrac{\log x}{\log 4} \\ \dfrac{\log 2x}{\log 2} &= \dfrac{\log x}{\log 2^2} \\ \dfrac{\log 2x}{\log 2} &= \dfrac{\log x}{2\log 2} \\ 2\log{2x} &= \log x\\ 2\log 2 + 2\log x &= \log x\\ \log x &= -2\log 2\\ \log x & = \log 2^{-2} \\ \log x & = \log \dfrac{1}{4}\end{align*}\] Therefore, \(x= \dfrac{1}{4}\). Substituting this value back into either of the original equations and we obtain that \(y=-1\). Therefore, the point of intersection is \(\left( \dfrac{1}{4}, -1\right)\).
We note first that \(x=a^y\) for all points on the curve. The midpoint of \(AB\) is given by \(\bigg(\dfrac{x_1+x_2}{2}\),\(\dfrac{y_1+y_2}{2}\bigg)\). Since we draw a horizontal line through the midpoint, \(y_3=\dfrac{y_1+y_2}{2}\), we have that \[\begin{align*} (x_3)^{2}&=(a^{y_3})^2\\ &=(a^\frac{y_1+y_2}{2})^2\\ &=(a^{y_1})(a^{y_2})\\ &=x_1x_2.\end{align*}\]
Given that the graph of the function passes through \((2,1)\), we know that \(a \neq 0\). We have \(1=a(2^r)\) and \(4=a(32^r)\). Since neither side of the first equation is 0, we can divide the second equation by the first to obtain \(4=\dfrac{32^r}{2^r} =\dfrac{2^r\cdot 16^r}{2^r}=16^r\). Therefore, \(r=\dfrac{1}{2}\).
Factoring both sides of the equation \(2^{x+3}+2^x=3^{y+2}-3^y\) gives \[\begin{align*} (2^3+1)2^x&=(3^2-1)3^y\\ 9 \cdot 2^{x}&=8 \cdot 3^{y}\\ 3^{2}\cdot 2^{x}&=2^{3}\cdot 3^{y}\\ 2^{x-3}&=3^{y-2}\end{align*}\] Since \(x\) and \(y\) are integers, and the only integer power of 2 that is also an integer power of 3 is the number \(1 = 2^0 = 3^0\), we have \(x = 3\) and \(y = 2\).
If \(f(x)=2^{4x-2}\), then \(f(x)\cdot f(1-x)=2^{4x-2}\cdot2^{4(1-x)-2}=2^{4x-2+4-4x-2}=2^0=1\).
Observe that the argument of both logarithms must be positive and so \(x>6\). Now \[\begin{align*} \log_5(x-2)+\log_5(x-6)&=2\\ \log_5((x-2)(x-6))&=2\\ (x-2)(x-6)&=25\\ x^2-8x-13 &=0 \\ x&=4\pm\sqrt{29}\end{align*}\] However, since \(x>6\), we have that \(x=4+\sqrt{29}\).
If \(a,b,c\) is a geometric sequence, then \(\dfrac{b}{a}=\dfrac{c}{b}\). It follows that \(\log_x\bigg(\dfrac{b}{a}\bigg)=\log_x\bigg(\dfrac{c}{b}\bigg)\) which implies \(\log_x b-\log_x a=\log_x c-\log_x b\). Therefore, the logarithms form an arithmetic sequence.
If \(\log_{x}a,\log_{x}b,\log_{x}c\) form an arithmetic sequence, then \[\begin{align*} \log_{x}b-\log_{x}a&=\log_{x}c-\log_{x}b\\ \log_{x}\left(\dfrac{b}{a}\right)&=\log_{x}\left(\dfrac{c}{b}\right)\\ \dfrac{b}{a}&=\dfrac{c}{b} &\text{(since the log function takes on each value only once)}\end{align*}\] Thus, \(a,b,c\) form a geometric sequence.
Using exponent rules and arithmetic, we manipulate the given equation: \[\begin{align*} 3^{x+2}+2^{x+2}+2^x & = 2^{x+5} + 3^x \\ 3^x 3^2 + 2^x 2^2 + 2^x & = 2^x 2^5 + 3^x \\ 9(3^x) + 4(2^x) + 2^x & = 32(2^x) + 3^x \\ 8(3^x) & = 27(2^x) \\ \dfrac{3^x}{2^x} & = \dfrac{27}{8} \\ \left(\dfrac{3}{2}\right)^x & = \left(\dfrac{3}{2}\right)^3\end{align*}\] Since the two expressions are equal and the bases are equal, then the exponents must be equal, and so \(x=3\).
Let \(a = \log_{10} x\). Then \((\log_{10}x)^{\log_{10}(\log_{10} x)} = 10\,000\) becomes \(a^{\log_{10}a} = 10^4\).
Taking the base \(10\) logarithm of both sides and using the fact that \(\log_{10}(a^b) = b\log_{10}a\), we obtain \((\log_{10} a)(\log_{10} a) = 4\) or \((\log_{10} a)^2 = 4\). Therefore, \(\log_{10} a = \pm 2\) and so \(\log_{10}(\log_{10} x) = \pm 2\).
If \(\log_{10}(\log_{10} x) = 2\), then \(\log_{10}x = 10^2 = 100\) and so \(x = 10^{100}\).
If \(\log_{10}(\log_{10} x) = -2\), then \(\log_{10}x = 10^{-2} = \frac{1}{100}\) and so \(x = 10^{1/100}\). Therefore, \(x = 10^{100}\) or \(x = 10^{1/100}\).
Note that \(x \neq 1\) since \(1\) cannot be the base of a logarithm. This tells us that \(\log x \neq 0\).
Using the fact that \(\log_a b = \dfrac{\log b}{\log a}\) and then using other logarithm laws, we obtain the following equivalent equations: \[\begin{align*} \log_4 x - \log_x 16 & = \dfrac{7}{6} - \log_x 8 \\ \dfrac{\log x}{\log 4} - \dfrac{\log 16}{\log x} & = \dfrac{7}{6} - \dfrac{\log 8}{\log x} & \text{(note that $x \neq 1$, so $\log x \neq 0$)}\\ \dfrac{\log x}{\log 4} & = \dfrac{7}{6} + \dfrac{\log 16 - \log 8}{\log x}\\ \dfrac{\log x}{\log(2^2)} & = \dfrac{7}{6} + \dfrac{\log(\frac{16}{8})}{\log x} \\ \dfrac{\log x}{2\log 2} & = \dfrac{7}{6} + \dfrac{\log 2}{\log x}\\ \dfrac{1}{2}\left(\dfrac{\log x}{\log 2}\right) & = \dfrac{7}{6} + \dfrac{\log 2}{\log x}\end{align*}\] Letting \(t = \dfrac{\log x}{\log 2} = \log_2 x\) and noting that \(t \neq 0\) since \(x \neq 1\), we obtain the following equations equivalent to the previous ones: \[\begin{align*} \dfrac{t}{2} & = \dfrac{7}{6} + \dfrac{1}{t} \\ 3t^2 & = 7t + 6 & \text{(multiplying both sides by $6t$)}\\ 3t^2 - 7t - 6 & = 0 \\ (3t+2)(t - 3) & = 0\end{align*}\] Therefore, the original equation is equivalent to \(t = -\frac{2}{3}\) or \(t= 3\).
Converting back to the variable \(x\), we obtain \(\log_2 x = -\frac{2}{3}\) or \(\log_2 x = 3\), which gives \(x = 2^{-2/3}\) or \(x = 2^3 = 8\).