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sim0nz12345 · Aug 3, 2007, 06:56 AM

Hi there, could you please tell me how to find these x values

Below is an enclosed attachment

Thanks

galactus · Aug 3, 2007, 10:41 AM

I will show you how to do the first one, then you do the second and post your attempt. I, or someone, will critique.

$2^{x}\cdot{2^{x+1}}=20$

When you multiply exponents, you add them.

x+x+1=2x+1, therefore,

and since $2^{2x+1}=2^{2x}\cdot{2}=2(2^{2})^{x}=2\cdot{4^{x}}$

$2\cdot{4^{x}}=20$

Divide by 2 and then ln of both sides:

$ln(4^{x})=ln(10)$

law of exponents:

$xln(4)=ln(10)$

$x=\frac{ln(10)}{ln(4)}$

I was just trying to show you the exponent laws. Learn them. They're easy once you practice a little. Very mechanical.

sim0nz12345 · Aug 3, 2007, 05:28 PM

Well for b), I guess I would bring the 5 over to the other side with the 5^x
But from there I don't know
Thanks for all the help so far