# Math Help - How to find a?

1. ## How to find a?

How do you find b? In:

-1/5 loga 3^-3 + 1/4 loga 3^1/4 = 53/320

I have gotten as far as working out:

loga 193/100 + loga 11/10 = 53/320 (at least I think I have done this part correctly)

but I am unsure where to go from here. I think its something to do with y = loga(x) which is a^y = x but I'm unsure how to take it forward other than trial and error.

2. Originally Posted by JQ2009
How do you find b? In:

-1/5 loga 3^-3 + 1/4 loga 3^1/4 = 53/320

I have gotten as far as working out:

loga 193/100 + loga 11/10 = 53/320 (at least I think I have done this part correctly)

but I am unsure where to go from here. I think its something to do with y = loga(x) which is a^y = x but I'm unsure how to take it forward other than trial and error.
You should be able to write this as:

$log_a(3^{\frac{1}{8}})-log_a(3^{\frac{3}{5}})=\frac{53}{320}$

Then use the property $log_a(x)-log_a(y)=log_a(\frac{x}{y})$ to write this as a single logarithm.

3. Should that not be:

log(x1 * x2) = logx1 + logx2 as the negatives cancel each other out (or does the - not go to the power with the 3?) and 1/16 rather than 1/8? (appologies if I have missed something).

Anyway doing it that way I got:

1822/100 is that correct?

4. Originally Posted by JQ2009
Should that not be:

log(x1 * x2) = logx1 + logx2 as the negatives cancel each other out Yes you're right! (or does the - not go to the power with the 3?) and 1/16 rather than 1/8? (appologies if I have missed something).

Anyway doing it that way I got:

1822/100 is that correct?
So this is what I end up with.

$log_a(3^{\frac{3}{5}+\frac{1}{16}})=\frac{53}{320}$

I don't see how you got 1822/100. I can't get a neat number like that. How did you proceed from what I wrote above?

5. I'm not sure now. I never wrote down my working for that part I punched it all into the calculator though I some how ended up with:

loga 2122/1000 = 53/320

so a^53/320 = 2122/1000

so a = root 53/320 of 2122/1000 (I'm not sure if thats expressed correctly).

But no matter what working I do now I cannot get 2122/1000, I must have made a mistake somewhere as doing what I though I'd done.

Also I don't understand how you are getting 1/8 rather than a 1/16, could you explain that please?

Anyway taking what you have up there I and doing what I did above I got 123.36

6. Originally Posted by JQ2009
I'm not sure now. I never wrote down my working for that part I punched it all into the calculator though I some how ended up with:

loga 2122/1000 = 53/320

so a^53/320 = 2122/1000

so a = root 53/320 of 2122/1000 (I'm not sure if thats expressed correctly).

But no matter what working I do now I cannot get 2122/1000, I must have made a mistake somewhere as doing what I though I'd done.

Also I don't understand how you are getting 1/8 rather than a 1/16, could you explain that please?

Anyway taking what you have up there I and doing what I did above I got 123.36
No, that was just another mistake on my part. I edited my last post.

$log_a(3^{\frac{3}{5}+\frac{1}{16}})=\frac{53}{320}$

If I add the fractions and pull out a calculator I get:

$log_a(3^{\frac{53}{80}})=log_a(3^{0.6625})$

So we can write:

$log_a(3^{0.6625})=\frac{53}{320}$

$a^{\frac{53}{320}}=3^{0.6625}$

$ln(a^{\frac{53}{320}})=ln(3^{0.6625})$

$\frac{53}{320}ln(a)=0.6625ln(3)$

$ln(a)=\frac{(320)(0.6625)ln(3)}{53}$

$a=e^{\frac{(320)(0.6625)ln(3)}{53}}$

lol, that's the best I can do to isolate $a$. I can't tell if there is a more simplistic way to get a 'neat' answer.