I have four problems I need help on:

5a^2/3*4a^3/2

(I believe it's multiplying, the copier messed up)

4a^5/3^3/2

log5(3x+1)=2

(It's log Base 5, as in the 5 there should be below the rest of the equation)

Log5x(x-1)=2

(This is base 10)

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- Aug 14th 2011, 09:18 AMLotr8808Exponents and Log Help!!
I have four problems I need help on:

5a^2/3*4a^3/2

(I believe it's multiplying, the copier messed up)

4a^5/3^3/2

log5(3x+1)=2

(It's log Base 5, as in the 5 there should be below the rest of the equation)

Log5x(x-1)=2

(This is base 10) - Aug 14th 2011, 09:22 AMSironRe: Exponents and Log Help!!
Do you have to simplify the exercice(s) with the exponents?

For the logarithmic equations, use: $\displaystyle \log_a(x)=y \Leftrightarrow a^y=x$

For example the second one:

$\displaystyle \log_5(3x+1)=2 \Leftrightarrow 5^2=3x+1 \Leftrightarrow ... $

(and offcourse if you want to do the solution check: $\displaystyle 3x+1>0$) - Aug 14th 2011, 09:25 AMwaqarhaiderRe: Exponents and Log Help!!
what you wants solution or simplification

- Aug 14th 2011, 09:27 AMLotr8808Re: Exponents and Log Help!!
I have to simplify and evaluate.

- Aug 14th 2011, 09:29 AMSironRe: Exponents and Log Help!!
Then you've to know the basic rules like for example $\displaystyle a^{x}\cdot a^{y}=a^{x+y}$, ... .

Can you continue with the hint I gave about the logarithmic equations?

Try to do some work by yourself, that's the best way to learn something. - Aug 14th 2011, 09:32 AMLotr8808Re: Exponents and Log Help!!
I did the first one very fast, but for the second one, how does the 5x affect the rule? Because I guess it would be 100 = 5x(x-1)

- Aug 14th 2011, 09:35 AMSironRe: Exponents and Log Help!!
If the equation is:

$\displaystyle \log[5x\cdot(x-1)]=2$ then indeed $\displaystyle 100=5x(x-1)\Leftrightarrow ... $.

What about the exercice with the exponents? - Aug 14th 2011, 09:35 AMLotr8808Re: Exponents and Log Help!!
I solved the second one by creating a quadratic and bringing everything to one side before factoring. I came up with 5(x-5)(x+4)

Now, I have to figure out the exponents. - Aug 14th 2011, 09:40 AMSironRe: Exponents and Log Help!!
Right! $\displaystyle x=-4$ and $\displaystyle x=5$ are indeed the solutions (don't forget to check them!)

But your first exercice is not clear, because you're not using brackets.

I guess you've to simplify:

$\displaystyle \frac{5a^2}{3}\cdot \frac{4a^3}{2}$

(to simplify, look at the 'basic rule' I gave in one of my previous posts) - Aug 14th 2011, 09:40 AMLotr8808Re: Exponents and Log Help!!
For the question: 4a^5/3^3^2, is there some way I can combine both exponents into one fraction?

- Aug 14th 2011, 09:43 AMLotr8808Re: Exponents and Log Help!!
Thanks for the rules, I will have to remember them! So, again, for the first one:

If I combine the two fractions:

I came up with

20a^10/6

After this, would the answer come out to 10a^10/3 - Aug 14th 2011, 09:47 AMSironRe: Exponents and Log Help!!
If the exercice is:

$\displaystyle 5a^{\frac{2}{3}}\cdot 4a^{\frac{3}{2}}=20a^{\frac{2}{3}+\frac{3}{2}}=20a ^{\frac{13}{6}}$

Now the second one, can you be more clear with using brackets? - Aug 14th 2011, 09:49 AMLotr8808Re: Exponents and Log Help!!
How does it come out to 13/6?

And, the question is:

$\displaystyle 4a^{\frac{5}{3}^{\frac{3}{2}$ - Aug 14th 2011, 10:03 AMSironRe: Exponents and Log Help!!
Because:

$\displaystyle \frac{3}{2}+\frac{2}{3}=\frac{9+4}{6}=\frac{13}{6}$

For the second one you can use this 'basic rule':

$\displaystyle (a^{x})^{y}=a^{x\cdot y}$ - Aug 14th 2011, 10:12 AMLotr8808Re: Exponents and Log Help!!
So, I got

$\displaystyle 32(a^{\frac{5}{2})$

And, for the top part, you cross-multiplied both ways and added them both?