How do I solve the following equation?

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

Thanks!

Guest Jun 18, 2017

#1**+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

CPhill
Jun 18, 2017

#2**+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

#3**+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