Solve 3^(2x+1)-10(3^X)+3=0, etc for X
- Thread starter Beans
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Have you not yet studied how to solve exponential or logarithmic equations...? Has your class at least covered logarithms, or are you needing help with that, too? :?:Beans said:
We're glad to help, but we need to know where, exactly, you're having difficulty. So please be specific. Thank you!
Eliz.
:arrow: P.S. Welcome to FreeMathHelp!
haha thanks for the welcome
Anyways, We studied them, but it was last year so i'm a little sketchy on the details.
this is sort of a group thing there called problem sets and are kinda crazy hard. Well, for the first problem we were thinking of changing them to all have 3 as the base but then your stuck with the -10. we changed the 3 to the other side of the problem and then divide by -10 but then your left with 3^(2x+1)+(3^x)=3/10 so i totally don't think thats right....
Anyways, We studied them, but it was last year so i'm a little sketchy on the details.
this is sort of a group thing there called problem sets and are kinda crazy hard. Well, for the first problem we were thinking of changing them to all have 3 as the base but then your stuck with the -10. we changed the 3 to the other side of the problem and then divide by -10 but then your left with 3^(2x+1)+(3^x)=3/10 so i totally don't think thats right....
For the first one:
\(\displaystyle \L\\3^{2x+1}-10\cdot{3^{x}}+3=0\)
Rewrite the first part as \(\displaystyle \L\\3(3^{x})^{2}\)
That way you have \(\displaystyle 3^{x}\) in all terms.
\(\displaystyle \L\\3(3^{x})^{2}-10\cdot{3^{x}}+3=0\)
Let \(\displaystyle \L\\u=3^{x}\)
This gives:
\(\displaystyle \L\\3u^{2}-10u+3=0\)
Now, solve the quadratic. Once you have 'u', you can use it to find x from
u=3^x.
\(\displaystyle \L\\3^{2x+1}-10\cdot{3^{x}}+3=0\)
Rewrite the first part as \(\displaystyle \L\\3(3^{x})^{2}\)
That way you have \(\displaystyle 3^{x}\) in all terms.
\(\displaystyle \L\\3(3^{x})^{2}-10\cdot{3^{x}}+3=0\)
Let \(\displaystyle \L\\u=3^{x}\)
This gives:
\(\displaystyle \L\\3u^{2}-10u+3=0\)
Now, solve the quadratic. Once you have 'u', you can use it to find x from
u=3^x.
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I'm not sure what the above means...? (Capitalization and punctuation can be very helpful in this regard.)Beans said:
:!: Note: I will be assuming, in what follows, that you mean "x" and "X" to actually be the same thing. (This is not mathematically standard, so please correct me if I am mistaken!) :!:
1) It might help to notice the following:
. . . . .3<sup>2x + 1</sup> = (3<sup>2x</sup>)(3<sup>1</sup>) = (3<sup>2x</sup>)(3) = 3 ((3<sup>x</sup>)<sup>2</sup>)
Then think about how you'd solve 3y<sup>2</sup> - 10y + 3 = 0.... :idea:
2) I will guess that you mean the logs to be "base eight". If so, then try thinking about the following:
. . . . .log<sub>8</sub>(x<sup>10</sup>) = 10 log<sub>8</sub>(x)
...and think about how you'd solve 3y<sup>2</sup> - 10y + 3 = 0. :wink:
Eliz.
galactus holy cow wow thank you so much, i never even thought about doing that for this problem. I'm a dork that makes it so easy! thanks!!!
AH i can so figure these out now thank you both of you! the U is the almighty letter of the day woots!! thank you thank you thank you both of you!! :lol:
AH i can so figure these out now thank you both of you! the U is the almighty letter of the day woots!! thank you thank you thank you both of you!! :lol: