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# HELLLLLPPPPP! due 2moro

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Find the sum of all the integer values of $m$ that make the following equation true: $\left(2^m3^5\right)^m 9^7=\dfrac{\left(256\cdot3^m\right)^m}{\left(\sqrt2\right)^{14}}$.

Jul 19, 2019

#1
+25237
+2

Find the sum of all the integer values of  $$m$$ that make the following equation true: $$\left(2^m3^5\right)^m 9^7=\dfrac{\left(256\cdot3^m\right)^m}{\left(\sqrt2\right)^{14}}$$ .

$$\begin{array}{|rcll|} \hline \left(2^m\cdot3^5\right)^m\cdot 9^7 &=& \dfrac{\left(256\cdot3^m\right)^m}{\left(\sqrt2\right)^{14}} \quad &| \quad \left(\sqrt2\right)^{14}=2^{\frac{14}{2}} = 2^7 \\ \left(2^m\cdot3^5\right)^m\cdot 9^7 &=& \dfrac{\left(256\cdot3^m\right)^m}{2^7} \quad &| \quad 256 = 2^8 \\ \left(2^m\cdot3^5\right)^m\cdot 9^7 &=& \dfrac{\left(2^8\cdot3^m\right)^m}{2^7} \quad &| \quad 9^7=(3^2)^7=3^{2\cdot 7}=3^{14} \\ \left(2^m\cdot3^5\right)^m\cdot 3^{14} &=& \dfrac{\left(2^8\cdot3^m\right)^m}{2^7} \quad &| \quad \cdot 2^7 \\ \left(2^m\cdot3^5\right)^m\cdot 3^{14}\cdot2^7 &=& \left(2^8\cdot3^m\right)^m \\ 2^{(m^2)}\cdot 3^{5m}\cdot 3^{14}\cdot2^7 &=& 2^{8m}\cdot3^{(m^2)} \\ \mathbf{2^{\overbrace{(m^2-8m+7)}^{=0}}} &=& \mathbf{3^{\overbrace{(m^2-5m-14)}^{=0}}} \quad |&\quad \text{The only solution is } 2^0 = 3^0 = 1 \\ 2^0 &=& 3^0 \\ \hline \end{array}$$

$$\text{Is there an equal m that sets m^2-8m +7 and m^2-5m-14 to zero at the same time?}$$

$$\begin{array}{|rcll|} \hline m^2-8m+7 &=& 0 \\\\ m &=& \dfrac{ 8\pm \sqrt{64-4\cdot 7} }{2} \\ m &=& \dfrac{ 8\pm 6 }{2} \\\\ m_1 &=& \dfrac{ 8+ 6 }{2} \\ \mathbf{m_1} &=& \mathbf{7} \\\\ m_2 &=& \dfrac{ 8- 6 }{2} \\ \mathbf{m_2} &=& \mathbf{1} \\ \hline \end{array} \begin{array}{|rcll|} \hline m^2-5m-14 &=& 0 \\\\ m &=& \dfrac{ 5\pm \sqrt{25-4\cdot(-14)} }{2} \\ m &=& \dfrac{ 5\pm 9 }{2} \\\\ m_1 &=& \dfrac{ 5+ 9 }{2} \\ \mathbf{m_1} &=& \mathbf{7} \\\\ m_2 &=& \dfrac{ 5- 9 }{2} \\ \mathbf{m_2} &=& \mathbf{-2} \\ \hline \end{array}$$

$$\text{So m is 7 and the sum of all the integer values of m is also \mathbf{7}} .$$

Jul 19, 2019

#1
+25237
+2

Find the sum of all the integer values of  $$m$$ that make the following equation true: $$\left(2^m3^5\right)^m 9^7=\dfrac{\left(256\cdot3^m\right)^m}{\left(\sqrt2\right)^{14}}$$ .

$$\begin{array}{|rcll|} \hline \left(2^m\cdot3^5\right)^m\cdot 9^7 &=& \dfrac{\left(256\cdot3^m\right)^m}{\left(\sqrt2\right)^{14}} \quad &| \quad \left(\sqrt2\right)^{14}=2^{\frac{14}{2}} = 2^7 \\ \left(2^m\cdot3^5\right)^m\cdot 9^7 &=& \dfrac{\left(256\cdot3^m\right)^m}{2^7} \quad &| \quad 256 = 2^8 \\ \left(2^m\cdot3^5\right)^m\cdot 9^7 &=& \dfrac{\left(2^8\cdot3^m\right)^m}{2^7} \quad &| \quad 9^7=(3^2)^7=3^{2\cdot 7}=3^{14} \\ \left(2^m\cdot3^5\right)^m\cdot 3^{14} &=& \dfrac{\left(2^8\cdot3^m\right)^m}{2^7} \quad &| \quad \cdot 2^7 \\ \left(2^m\cdot3^5\right)^m\cdot 3^{14}\cdot2^7 &=& \left(2^8\cdot3^m\right)^m \\ 2^{(m^2)}\cdot 3^{5m}\cdot 3^{14}\cdot2^7 &=& 2^{8m}\cdot3^{(m^2)} \\ \mathbf{2^{\overbrace{(m^2-8m+7)}^{=0}}} &=& \mathbf{3^{\overbrace{(m^2-5m-14)}^{=0}}} \quad |&\quad \text{The only solution is } 2^0 = 3^0 = 1 \\ 2^0 &=& 3^0 \\ \hline \end{array}$$

$$\text{Is there an equal m that sets m^2-8m +7 and m^2-5m-14 to zero at the same time?}$$

$$\begin{array}{|rcll|} \hline m^2-8m+7 &=& 0 \\\\ m &=& \dfrac{ 8\pm \sqrt{64-4\cdot 7} }{2} \\ m &=& \dfrac{ 8\pm 6 }{2} \\\\ m_1 &=& \dfrac{ 8+ 6 }{2} \\ \mathbf{m_1} &=& \mathbf{7} \\\\ m_2 &=& \dfrac{ 8- 6 }{2} \\ \mathbf{m_2} &=& \mathbf{1} \\ \hline \end{array} \begin{array}{|rcll|} \hline m^2-5m-14 &=& 0 \\\\ m &=& \dfrac{ 5\pm \sqrt{25-4\cdot(-14)} }{2} \\ m &=& \dfrac{ 5\pm 9 }{2} \\\\ m_1 &=& \dfrac{ 5+ 9 }{2} \\ \mathbf{m_1} &=& \mathbf{7} \\\\ m_2 &=& \dfrac{ 5- 9 }{2} \\ \mathbf{m_2} &=& \mathbf{-2} \\ \hline \end{array}$$

$$\text{So m is 7 and the sum of all the integer values of m is also \mathbf{7}} .$$

heureka Jul 19, 2019