+2446 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. |
| \(x=-9\) | |
