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Toaster · Aug 8, 2009, 06:44 PM

Hey guys,

I took a year off and now I'm back in the books.

Now it might be because I've been doing nothing but math today and my brain is fried, but I can't figure out these two problems. Any help is greatly appreciated!

Solve for the variable:

a) logy81=4/3
so I've rewritten it as:
y^4/3= 81
y=?
and that's where I'm at. I have no clue what the base of y is or how to find it.

b)log8 Y=2/3
8^2/3=y
y=?

How am I suppose to write the value for y leave it as 8^2/3?

I hope that wasn't to confusing but thanks for taking a look.

Trevor

galactus · Aug 9, 2009, 03:19 AM

$log_{y}(81)=\frac{4}{3}$

What this is asking is what number raised to the 4/3 power is equal to 81?

$y^{\frac{4}{3}}=81$

Try the change of base formula: $\frac{log(81)}{log(y)}=\frac{4}{3}$

$\frac{4log(3)}{3log(3)}=\frac{4}{3}$

In other words, what is 3^3 equal?

Unknown008 · Aug 9, 2009, 05:45 AM

I don't know if you'll understand my method, but here it is.

$log_y (81) = \frac43$

$y^{\frac43} = 81$

But what is 81? It's also $3^4$

So,

$y^{\frac43} = 3^4$

Then 'cube' both sides, then take the forth root of both sides.

$y^{\frac43 \times 3} = 3^{4\times 3 }$

$y^4 = 3^{12}$

$y^{4 \times \frac14} = 3^{12\times \frac14}$

$y= 3^{3}$

Or simply take the power of 3/4 on both sides.

$y^{\frac43 \times \frac34} = 3^{4\times\frac34}$

$y= 3^{3}$

And for the next one, $8 = 2^3$

Toaster · Aug 9, 2009, 06:05 AM

Wonderful thanks guys! The problem was I had just forgotten which methods to use, so this help a lot.

@ unknown088, I get what you've done, but what is this formula called?

Thanks again guys!

Unknown008 · Aug 9, 2009, 06:09 AM

Huh? Which formula?

These are called logarithmic equations if that was the word you're looking for...

You make use of the power law...

Toaster · Aug 9, 2009, 08:27 AM

Ah k I literally just learned the power law so that makes sense now.

Thanks again!

Toaster · Aug 9, 2009, 05:35 PM

[math]x{\small_1}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/máth]

[math]\normalsize\log_{2}(2m+4)-\log_2(m-1)=3[/máth]

(lets see if this actually works)

galactus · Aug 9, 2009, 05:46 PM

Originally Posted by Toaster

$x{\small_1}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\log_{2}(2m+4)-\log_2(m-1)=3$

(lets see if this actually works)

What is that little thing above the a in [math]? That is why it won't display correctly.

I got rid of it and now it works.

Toaster · Aug 9, 2009, 05:48 PM

Ah thanks galactus! I'll use it next time, by the way I posted a new question if you wouldn't mind taking a look.

Unknown008 · Aug 10, 2009, 08:36 AM

LOL! That was the symbol that RickJ posted in his post so that you could see what was the code tags! He specified in his post that you should replace the a symbol by a for the code to work.

EDIT: that was Capuchin, not RickJ, sorry :o

Toaster · Aug 10, 2009, 09:02 AM

Yea what can I say I was extremely tired/didn't read anything just copied and pasted the tags.