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

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?

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

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!

Huh? Which formula?

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

You make use of the power law...

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

Thanks again!

[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)

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

\normalsize\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.

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.

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

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