g^x+2=(1/27)^1-x

Question

Solve for x in this problem

Answer 1

Starting with g^(x+2)=(1/27)^(1-x)

 

we can take logs of both sides giving (x+2)log(g)=(1-x)log(1/27) now the log(1/27)=log(1)-log(27)=-log(27) since the log(1)=0

 

So now we have (x+2)log(g)=-(1-x)log(27)=(x-1)log(27) expanding and getting the terms with x's in them all on one side gives

 

x(log(g)-log(27))=-log(27)-2log(g) now we can solve for x which is

 

x=(3log(3)+2log(g))/(3log(3)-log(g)) where I have used log(27)=log(3³)=3log(3)

 

Regards

Jim

Answer 2

Hi Chalyce,

 

Remember that solving for x means we want to rearrange the terms in the equation to find "x=[something]". In this case, since we have another variable -- g -- to deal with, our "something" won't be a single number, it'll be an expression with g in it.

 

I'm going to assume, even though you haven't put the parentheses in, that your question is actually g^x+2 = (1/27)^1-x -- so that the addition operations are actually in the powers. Is that correct? If not, we'll need to redo the problem; the answers will be different.

 

This is a pretty complicated problem! It's actually taking me a minute to go through it :P good thing you asked for help!

 

Let's start by looking at those x's in the powers. They're driving me crazy! None of my usual PEMDAS operations can get them out of there, or combine them. Not even logs can help us yet, because we have TWO instances of x being in the exponent -- this'll make trouble. Yeesh. What to do? Well, we can actually rewrite each side of the equation using our properties of exponents. 

 

• (a^m)(aⁿ) = a^(m+n)
• (a^m)ⁿ = a^mn
• (a/b)^m = (a^m)/(b^m)
• a^m/aⁿ = a^m-n

 

Okay. Apply the right side of the first property to the right side of the equation: our base, a, is 1/27, and it's to the power of 1-x, so our m is 1 and our n is -x (yes, you can make it negative and even make it a variable! I love algebra.). So since we have a=1/27, m=1, and n=-x, we can make the right side of the equation follow the structure of the left side of the property, and write (1/27)¹·(1/27)^-x. Whew!

 

We can do the exact. same. thing. with the left side of the equation: our base, a, is g, and it's to the power of x+2, so our m is x and our n is 2. So a=g, m=x, n=2; we can make the left side of the equation follow the structure of the left side of the property, and write g^x·g². Tired yet? ;)

 

Just to make sure we don't forget about the structure of the equation, let's write out what we've got now:

g^x·g² = (1/27)¹ · (1/27)^-x_.

 

Well, what can we simplify next? Right away, I see that we have (1/27)¹. That's just (1/27), yeah? Cool. What else? Ummm... We have (1/27)^-x. Since -x can be rewritten as -1·x, or x·-1, we can rewrite this bit as (1/27)^x·-1. Then we can use the second property to rewrite this as ((1/27)^x)^-1. Then, since PEMDAS tells us to do the inside parentheses first, we can use the third property to rewrite ((1/27)^x)^-1) as (1^x/27^x)^-1.

 

Now, almost done with this term -- we know that no matter how many times you multiply 1 by itself, 1² or 1⁰ or 1^10000000 or 1^x, it will still be 1, so rewrite again as simply (1/27^x)^-1. Then, knowing that putting something to the power of -1 means to find the reciprocal, flip this fraction and get (27^x)/1, or just 27^x. WHEW. That was a lot! If you need clarification on any of this section, please ask :)

 

Okay. Big picture time -- where were we? What did we just do? We rewrote (1/27)¹ as (1/27), and then we rewrote (1/27)^-x a bunch of times until it became 27^x. So let's put those rewrites into our equation:

 

g^x·g² = (1/27)¹ · (1/27)^-x       ----->>      g^x·g² = (1/27) · 27^x.

 

Okay... Now we have a g^x on the one side, and a 27^x on the other, and they're simple enough if we switch one of those terms to the other side and combine them, they'll turn into something nice. Let's try it. Divide both sides by 27^x. While we're at it, we know that g² has nothing to do with x's, so let's switch it to the other side by division as well:

 

(g^x·g²) ·(1/g²) ·(1/27^x) = ((1/27) · 27^x) · (1/27^x) · (1/g²)

 

(I ended up writing the divisions as multiplications-by-reciprocals for clarity -- I'm sure you remember they're the same thing.) 

 

This turns into (g^x·g²) ·(1/g²) ·(1/27^x) = ((1/27) · 27^x) · (1/27^x) · (1/g²), which then simplifies to g^x/27^x = (1/27)/g², or just 1/(27g²).

 

Look at that! Our left side is just x-bits, and our right side is just non-x-bits. Awesome. Let's simplify that left side, and see if we can reach some kind of x= thing.

 

Use the fourth property to rewrite the left side as (g/27)^x. Excellent. Now we just have ONE x in the whole equation, and NOW we can use logs to make it nice.

 

So let's take what we have : (g/27)^x = 1/(27g²), and turn it into a log function. Remember that the definition of logs means we can convert between:

• b^m = a  <-->  log_b(a) = m.

 

Our equation is a LITTLE more complex than that -- but really, our a is just the 1/(27g²), and the b is g/27, and the m is x. So we can write: log_(g/27)(1/(27g²)) = x. Whoa! We got an "x="!!! :D :D

 

I think it can be prettier, though. That log expression has too many fractions in it for my liking. Let's see if we can't clean it up some.

 

Properties of logarithms include:

• log_a(p/q) = log_ap – log_aq

Alright, so use this to convert our fractional log expression (with numerator p=1 and denominator q=27g2) to a subtraction log expression, which is log_(g/27)(1)-log_(g/27)(27g2). As for that first term -- one would read that first term as, "(g/27) to some power is 1." You know that -- it's 0! Anything to the 0 power is 1. So replace that entire first term with 0. Almost done!!!!

 

We now have x = -log_(g/27)(27g2). I'm gonna go ahead here and tell you that you can't do anything else -- trying to take the fraction out of the log base is useless, and you COULD separate it into two log operations added together, but it's more elegant this way. That's it :) 

 

Took me about an hour to write this out, but I promise it won't take that long to solve once you get the hang of it! Keep a sheet of logarithm and exponent properties nearby, use them whenever it looks useful, and you'll be flying through problems in no time :)

 

As always, if you have any questions, feel free to ask!

 

Brianna