+0  
 
0
675
3
avatar

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

Guest Jul 14, 2015

Best Answer 

 #3
avatar+27228 
+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
 #1
avatar+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
avatar+20680 
+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
avatar+27228 
+13
Best Answer

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

33 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.