$${{\mathtt{5}}}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)} = {\mathtt{104}} \Rightarrow {{\mathtt{5}}}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)} = {\mathtt{104}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)} \Rightarrow {{\mathtt{5}}}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)} = {\mathtt{104}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}$$

Guest Aug 15, 2015

#1**+5 **

If you want to solve the equation look at http://web2.0calc.com/questions/5-x-1-5-2-x-2-104

Evaluating the left-hand side of the equation at x = 5 does not result in 104.

$${{\mathtt{5}}}^{\left({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{2}}\right)} = {\mathtt{15\,665}}$$

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Alan Aug 15, 2015

#3**+10 **

Best Answer

To get the answer of 5, change the first '5' into a '2':

2^{x + 1} + 5·2^{x - 2} = 104

If you graph the function y = 2^{x + 1} + 5·2^{x - 2} - 104, you will get the solution x = 5.

geno3141 Aug 15, 2015

#4**+5 **

Well spotted geno!

2^(x+1) + 5*2^(x-2) = 104

2^x*(2 + 5/4) = 104

2^x*13/4 = 104

2^x = 104*4/13

2^x = 32

2^x = 2^5

x = 5

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Alan Aug 16, 2015

#5**+5 **

I understand both answers but I do not understand this statement Geno

To get the answer of 5, change the first '5' into a '2'

(**Sorry I get you now** - how on Earth did you figure that out!)

What does that mean and what has it got to do with your graphical solution ?

Your algebra is really neat Alan

Geno has said that if

2^{x + 1} + 5·2^{x - 2} = 104

then

2^{x + 1} + 5·2^{x - 2} - 104 =0

If we graph

y=2^{x + 1} + 5·2^{x - 2} - 104

and we graph

y=0 (this is just the x axis)

then, where the two graphs cross on the x axis will be the solution.

This is a technique that we use all the time but I think many students (even those who use it ) do not really understand what they are doing. :)

Here is the graph.

Melody Aug 16, 2015