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How do I solve the following equation?

 (2^(3x)*8^(x+3))/16^(x-1)=2^((x-1)/2)

 

Thanks!

 
Guest Jun 18, 2017
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 #1
avatar+74645 
+1

 

(2^(3x) * 8^(x + 3) / 16^(x - 1)   =  2^( (x - 1) /2 )  

 

We can  write this as

 

 2^(3x) *  (2^3)^(x + 3)  / (2^4)^(x -1) = 2^( ( x - 1) / 2)

 

 2^(3x) * (2)^(3x + 9) / 2^(4x - 4)  = 2^( ( x - 1) /2)

 

2^(3x) * (2)^(3x + 9) = 2^( (x - 1) / 2) * 2^(4x - 4)

 

2^( 3x + 3x + 9)   =  2^( (x - 1) /2 +  (8x - 8) /2 )

 

2^ (6x + 9) =  2^( ( x - 1 + 8x - 8) /2)

 

2^( 6x + 9 )  = 2^( (9x - 9) / 2 )

 

Since the bases are the same we can solve for the exponents

 

6x + 9  = (9x - 9) / 2    multiply both sides by 2

 

12x + 18   =  9x - 9     subtract 9x , 18 from both sides

 

3x = -27     divide both sides by 3

 

x  = -9

 

 

cool cool cool

 
CPhill  Jun 18, 2017
 #2
avatar+477 
+1

Let me try to do this. If I am not mistaken, this is the original equation:

 

\(2^{3x}*\frac{8^{x+3}}{16^{x-1}}=2^{\frac{x-1}{2}}\)

 

\(2^{3x}*\frac{8^{x+3}}{16^{x-1}}=2^{\frac{x-1}{2}}\)Our first step, I think, is to get rid of the fraction. I'm going to use the rule that \(\frac{1}{a^b}=a^{-b}\)
\(\frac{1}{16^{x-1}}=16^{-(x-1)}=16^{-x+1}\)Doing this puts allows me to take this out of the fraction. Therefore, I am going to rewrite the current equation
\(2^{3x}*8^{x+3}*16^{-x+1}=2^{\frac{x-1}{2}}\)Now, convert \(8^{x+3}\)into a form where it will be in base 2. Luckily for us, all these numbers can be in that form. 
\(8^{x+3}=(2^3)^{x+3}\)Okay, let's insert that back into the equation.
\(2^{3x}*(2^3)^{x+3}*16^{-x+1}=2^{\frac{x-1}{2}}\)Now, convert \(16^{-x+1}\) into base 2, as well. 
\(16^{-x+1}=(2^4)^{-x+1}\)Insert that into the original equation again, too.
\(2^{3x}*(2^3)^{x+3}*(2^4)^{-x+1}=2^{\frac{x-1}{2}}\)Now, I'll apply a rule on the term \((2^3)^{x+3}\) that says that \((a^b)^c=a^{b*c}\). Let's use it.
\((2^3)^{x+3}=2^{3(x+3)}\)Insert it into the original equation.
\(2^{3x}*2^{3(x+3)}*(2^4)^{-x+1}=2^{\frac{x-1}{2}}\)Now, I'll use another power rule that says that \(a^b*a^c=a^{b+c}\). I'll utilize this for \(2^{3x}*2^{3(x+3)}\)
\(2^{3x}*2^{3(x+3)}=2^{3x+3(x+3)}\)Reinsert this back into the equation.
\(2^{3x+3(x+3)}*(2^4)^{-x+1}=2^{\frac{x-1}{2}}\)Okay, now the only term left is \((2^4)^{-x+1}\). Just like before, we'll use an exponent rule that says that \((a^b)^c=a^{b*c}\).
\((2^4)^{-x+1}=2^{4(-x+1)}\)Insert this back into the equation again.
\(2^{3x+3(x+3)}*2^{4(-x+1)}=2^{\frac{x-1}{2}}\)Yet again, we'll utilize the same rule as before that says that \(a^b*a^c=a^{b+c}\).
\(2^{3x+3(x+3)}*2^{4(-x+1)}=2^{3x+3(x+3)+4(-x+1)}\)Reinsert this into the equation again.
\(2^{3x+3(x+3)+4(-x+1)}=2^{\frac{x-1}{2}}\)Now, we'll use another rule that says that\(a^{f(x)}=a^{g(x)},\text{then}\hspace{1mm}f(x)=g(x)\). This will reduce the equation to simply two-sided equation without exponents.
\(3x+3(x+3)+4(-x+1)=\frac{x-1}{2}\)To clean this up, let's use the distribute property.
\(3x+3x+9-4x+4=\frac{x-1}{2}\)Combine like terms on the left hand side of the equation.
\(2x+13=\frac{x-1}{2}\)Multiply both sides by 2 to get rid of the pesky fraction.
\(4x+26=x-1\)Subtract x on both sides.
\(3x+26=-1\)Subtract 26 on both sides
\(3x=-27\)Divide by 3 on both sides to finally isolate x.
\(x=-9\) 
  
 
TheXSquaredFactor  Jun 18, 2017
edited by TheXSquaredFactor  Jun 18, 2017
 #3
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+1

OK, let everybody chime in !!.

 

Solve for x:
2^(2 x + 13) = 2^((x - 1)/2)

Take the natural logarithm of both sides and use the identity log(a^b) = b log(a):
log(2) (2 x + 13) = 1/2 log(2) (x - 1)

Divide both sides by log(2):
2 x + 13 = (x - 1)/2

Expand out terms of the right hand side:
2 x + 13 = x/2 - 1/2

Subtract x/2 + 13 from both sides:
(3 x)/2 = -27/2

Multiply both sides by 2/3:
Answer: | x = -9

 
Guest Jun 18, 2017

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