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# solve for x: (3^(x+2))*(5^(x-1))=15^(2x)

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solve for x: (3^(x+2))*(5^(x-1))=15^(2x)

Guest Jul 14, 2015

#3
+26412
+13

This could be made a little less "loggy" as follows:

$$\\3^{x+2}5^{x-1}=15^{2x}\\\\3^x3^25^x5^{-1}=15^{2x}\\\\(3\times5)^x\times\frac{9}{5}=15^{2x}\\\\15^x\times\frac{9}{5}=15^{2x}\\\\15^{2x-x}=\frac{9}{5}\\\\15^x=\frac{9}{5}\\\\x\times\log15=\log{\frac{9}{5}}\\\\ x=\frac{\log1.8}{\log15}$$

$${\mathtt{x}} = {\frac{{log}_{10}\left({\mathtt{1.8}}\right)}{{log}_{10}\left({\mathtt{15}}\right)}} \Rightarrow {\mathtt{x}} = {\mathtt{0.217\: \!051\: \!613\: \!246\: \!638\: \!7}}$$

heureka beat me to it!

Alan  Jul 14, 2015
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#1
+423
+13

$$\left({{\mathtt{3}}}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}\right){\mathtt{\,\times\,}}\left({{\mathtt{5}}}^{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}\right) = {{\mathtt{15}}}^{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}$$

We got a lot of messyness in the powers here. We can split them into more individual numbers to seperate off the constants and variables.

$${{\mathtt{3}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{-{\mathtt{1}}} = {{\left({{\mathtt{15}}}^{{\mathtt{2}}}\right)}}^{{\mathtt{x}}}$$

$${\mathtt{9}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{x}}} = {{\mathtt{225}}}^{{\mathtt{x}}}$$

$$\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{x}}} = {{\mathtt{225}}}^{{\mathtt{x}}}$$

Now we can put everything into logarithm form. Remember:

Log(a^b) = b * log(a)

Log(a*b) = log(a) + log(b)

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{5}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{225}}\right)$$

Let's move over all the 'x' multiples over, so we can handle them together.

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{225}}\right){\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{5}}\right){\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right)$$

Factorise it a little, so we can now split the "x" from the rest of the formula

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}\left({log}_{10}\left({\mathtt{225}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{5}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{3}}\right)\right)$$

Simplify the logarithms and re-arrange the forumula

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\frac{\left({\frac{{\mathtt{225}}}{{\mathtt{5}}}}\right)}{{\mathtt{3}}}}\right)$$

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{15}}\right)$$

$${\frac{{log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right)}{{log}_{10}\left({\mathtt{15}}\right)}} = {\mathtt{x}}$$

$${\mathtt{x}} = {\mathtt{0.217\: \!051\: \!613\: \!246\: \!638\: \!7}}$$

Sir-Emo-Chappington  Jul 14, 2015
#2
+18843
+13

solve for x: (3^(x+2))*(5^(x-1))=15^(2x)

$$\small{\text{ \begin{array}{rcl} 3^{x+2} \cdot 5^{x-1} &=& 15^{2x}\\ 3^x\cdot 3^2 \cdot 5^x \cdot 5^{-1} &=& 15^{2x}\\ 3^x\cdot 5^x \cdot 3^2 \cdot 5^{-1} &=& 15^{2x}\\ ( 3\cdot 5)^x \cdot 3^2 \cdot 5^{-1} &=& 15^{2x}\\ 15^x \cdot \dfrac{9}{5} &=& 15^{2x}\\ 15^x \cdot 1.8 &=& 15^{2x}\\ 15^x \cdot 1.8 &=& 15^{x+x}\\ 15^x \cdot 1.8 &=& 15^x\cdot 15^x \qquad | \qquad : 15^x\\ 1.8 &=& 15^x\\ 15^x &=& 1.8 \qquad | \qquad \ln{()}\\ \ln{( 15^x )} &=& \ln{ (1.8) } \\ x\cdot \ln{( 15 )} &=& \ln{ (1.8) } \\\\ x &=& \dfrac{ \ln{ (1.8) } } { \ln{( 15 )} } \\ \\ x &=& \dfrac{ 0.58778666490 } { 2.70805020110 } \\\\ \mathbf{x} & \mathbf{=} & \mathbf{0.21705161325} \end{array} }}$$

heureka  Jul 14, 2015
#3
+26412
+13

This could be made a little less "loggy" as follows:

$$\\3^{x+2}5^{x-1}=15^{2x}\\\\3^x3^25^x5^{-1}=15^{2x}\\\\(3\times5)^x\times\frac{9}{5}=15^{2x}\\\\15^x\times\frac{9}{5}=15^{2x}\\\\15^{2x-x}=\frac{9}{5}\\\\15^x=\frac{9}{5}\\\\x\times\log15=\log{\frac{9}{5}}\\\\ x=\frac{\log1.8}{\log15}$$

$${\mathtt{x}} = {\frac{{log}_{10}\left({\mathtt{1.8}}\right)}{{log}_{10}\left({\mathtt{15}}\right)}} \Rightarrow {\mathtt{x}} = {\mathtt{0.217\: \!051\: \!613\: \!246\: \!638\: \!7}}$$

heureka beat me to it!

Alan  Jul 14, 2015

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