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817
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avatar+8 

Can some explain to me how is this solved: 2^(1-4x) = 32

What's the process to finding x?

The calculator outputs:$${\mathtt{x}} = {\mathtt{\,-\,}}\left({\frac{\left({ln}{\left({\mathtt{32}}\right)}{\mathtt{\,-\,}}{ln}{\left({\mathtt{2}}\right)}\right)}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{2}}\right)}\right)}}\right)$$

 Jul 28, 2015

Best Answer 

 #1
avatar+128089 
+10

2^(1 -4x) = 32       notice that 32 = 2^5 ....so we have....

 

2^(1-4x)  = 2^5       since we have bases the same, we can solve for the exponents......this gives....

 

1 - 4x   =  5       subtract 1 from both sides

 

-4x = 4     divide by -4 on both sides

 

x = -1

 

 

  

 Jul 28, 2015
 #1
avatar+128089 
+10
Best Answer

2^(1 -4x) = 32       notice that 32 = 2^5 ....so we have....

 

2^(1-4x)  = 2^5       since we have bases the same, we can solve for the exponents......this gives....

 

1 - 4x   =  5       subtract 1 from both sides

 

-4x = 4     divide by -4 on both sides

 

x = -1

 

 

  

CPhill Jul 28, 2015
 #2
avatar+8 
+5

whoaw you're a pro man, thanks 

 Jul 28, 2015
 #3
avatar+128089 
0

No prob

 

 

  

 Jul 28, 2015

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