What is the solution to the equation 64^(4-x)=4^(2x)? They want me to use complete sentences explaining how I did this so could you give me the steps on how to solve it? It has to be in approximate decimal form.
hi
hey. you wanna solve? lol. because I can't.
What grade are you?
11th
I thought you were 19...
it's a real long story, man. I had to drop out because of family and now im taking it online.
oh....sorry for bringing it up >.>
nah it's okay. you aren't gonna solve, are you?
Whats the ^ mean?
Powers?
yeah, it's an exponent.
I know more than you :3 that's sad... lol
I technically already knew what it was...I was just making sure ;)
oh okay, then by all means, proceed
Idk...usually being called a turd doesnt inspire one to do work for ya ;)
you don't even know how to do it, do you? lmao
I probably do...somewhere in my head...but you should probably ask someone who hasnt been up for over a day ;)
well I asked and you were the only who replied soo... lol. I took the ACT today so I have no more brain left for more math D:
Get some rest then :)
Psshh I wish. I just have this one assignment left and im trying to get it done
Is it a " last minute " one ;P
:/ yes. -____________-
Welcome to the Club
64 = 4 x 4 x 4 = 4^3 Replace 64 with 4^3. Simplify. Since the bases are the same on both sides of the equation, the exponents must be the same. Equate the exponents and solve for x.
Wow...nice was of putting it ranga
Thank you, ranga, for helping me on this :) I appreciate it!
You greens know your stuff :)
Thank you @ShadowLegendX. You are wleocme @megangray32.
\[\Large 64^{4-x} = 4^{2x}\]\[\Large (4^{3})^{4-x} = 4^{2x}\]\[\Large 4^{12 - 3x} = 4^{2x}\]Bases are the same. Therefore, exponents must be the same. \[\Large 12 - 3x = 2x\]Solve for x.
12/5, right?
Yes, but they want the answer in decimal.
oh yeah, it'd be 2.4 then?
Yes.
okay thank you again
Glad to be able to help.
*walks through question* Don't mind me, just stalking lalala~
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