Solve for x: A Frustrating Challenge!

[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

 

[T@P]

test for good formating...

2^x

 

[footprints]

 

[T@P]

hmm sorry i guess I am 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 I am not misleading you or that I am too confusing..

 

[footprints]

simply re-write 2^(x+1) as (2^x) * 2

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

 

[T@P]

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

 

[T@P]

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

 

[footprints]

actually how do you write things nicely here?

Its called latex. To end a sentence in a latex form u have to type [/tex] and to start u have to type $$Click on quote to see how on the bottom right hand corner of a post to see how people use latex$$

 

[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 $$2^{2x} \cdot 2 = 3^2$$

 

[T@P]

thanks a lot 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[/tex]

so by taking both sides base 2, you get that x = log(2) base 3 (i don't 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...

 

[T@P]

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

 

[T@P]

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

 

[footprints]

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

x = log(2) base 3 (i don't know the latex for this)

$$x = log_2{3}$$

 

[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)

 

[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

 

[Muzza]

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

 

[footprints]

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

 

[T@P]

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

 

[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.