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# Help with Exponential Equations

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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
#1
+88899
+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
+2190
+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
+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: