1. Oct 27, 2004

footprints

Solve for x.
$$2^{x+1} + 2^x = 9$$

$$lg(x-8) + lg(\frac{9}{2}) = 1+ lg(\frac{x}{4})$$

I can't get the answers :grumpy:

2. Oct 27, 2004

T@P

test for good formating...
Code (Text):
2^x

3. Oct 27, 2004

footprints

4. Oct 27, 2004

T@P

hmm sorry i guess im not smart enough to use it.

anyway to answer your question, for the first simply re-write 2^(x+1) as (2^x) * 2, and then factor out 2^x.

this leaves (2^x) * (2+1) = 9
or 2^x = 3 from which x = Log (base 2) 3
for the second use the basic law the ln(x) + ln(y) = ln(x *y)
I hope im not misleading you or that im too confusing..

5. Oct 27, 2004

footprints

Do u mean $$2^{2x}$$?

6. Oct 27, 2004

T@P

no. for example 3^5 = 3^4 *3 (adding exponents) so it would be (2) * (2^x)

7. Oct 27, 2004

T@P

actually how do you write things nicely here? im making everyone confused with my poor notation and my test failed :(

8. Oct 27, 2004

Its called latex. To end a sentence in a latex form u have to type [/tex] and to start u have to type $$Click on quote to see how on the bottom right hand corner of a post to see how people use latex 9. Oct 27, 2004 footprints I don't really get what u mean from this sentence onwards (2^x) * (2+1) = 9 but i'll just show u what i can do for this equation so far. I stuck after i get this $$2^{2x} \cdot 2 = 3^2$$ 10. Oct 27, 2004 T@P thanks alot for the LaTex help :) anyway you agree that $$2^3 \cdot 2 = 2^4$$ im not trying to sound too obvious but this is essentially adding the exponents. This is simply $$2^{x} \cdot 2 + 2^{x} = 9$$ or by pulling out $$2^{x}$$ it is $$2^{x} \cdot (3 + 1) = 9$$ $$2^{x} = 3$$ generally, log (x^y) = ylog(x) this is basic rule that is always true regardless of the base or anything. in our case, by taking the logarithm of both sides to get linear equation in the form [tex} a \cdot x = b$$

so by taking both sides base 2, you get that x = log(2) base 3 (i dont know the latex for this)

anyway another basic rule i used here was that
log (x) base x = 1
this can be intuitively seen as the answer to the question, "what power do i raise x to to get x".

the answer is one, since $$x^1 = x$$

hope that helps...

i also hope i didnt mess up the latex...

11. Oct 27, 2004

T@P

aside from a few typos pretty good for a first try if i may say so myself :)

12. Oct 27, 2004

T@P

sorry in one equation i miswrote it it actually is $$2^x \cdot (2 + 1) = 9$$ not (3 + 1) sorry

13. Oct 27, 2004

footprints

Oops i made a mistake it isn't $$2^{2x} \cdot 2 = 3^2$$.
$$x = log_2{3}$$

Last edited: Oct 28, 2004
14. Oct 27, 2004

vsage

I'm not sure what "lg" means so I will assume that is a base 10 logarithm. Consider What "1" is expressed as a base 10 logarithm. You know that log(a) + log(b) = log(ab) so I would start there in solving the second one. The rest should follow fairly easily.

Edit they made a typo above. It should be 2^x(2+1)

Last edited by a moderator: Oct 28, 2004
15. Oct 28, 2004

footprints

So the correct equation is $$2^x (2+1) = 3^2$$
Then i take $$log_2$$ of both sides.
$$x log_2{2} + log_2{(2+1)} = 2 log_2{3}$$
Right?
Got a feeling i did somethin wrong

16. Oct 28, 2004

Muzza

Why don't you just divide both sides with 3?

17. Oct 28, 2004

footprints

Oh never mind. I just solved both. Thanks for sharing.

18. Oct 28, 2004

T@P

yes you did do it right on the previous page. Hope i helped you out somewhat :)

19. Oct 28, 2004

footprints

Oh u did help in some way. Its just that i did it another way. I let $$y = 2^x$$ then solve from there.