• Jul 27th 2010, 12:42 PM
garougt
I'm basically done all the questions that I needed to do, yet this one I can't seem to find how to answer for some reason.

3(2)^x = 4^(x + 1)

To find x, I try to put it to logarithmic function,

3*(x)(Log 2) = (x + 1)(Log 4)

My question is, is that where the 3 goes? It multiplies (x)(Log 2)? Doing it this way I end up with,

0.9x - 0.6x = 0.6
x = 2

The correct answer however is -0.42

I really need help what to do with that 3, as I've read the book plenty now and I still don't know how this works! Please help!

There's also another question similar to this, and of course I can't answer it either.

5(4)^2x = 4(2)^6x

Any help is greatly appreciated!
• Jul 27th 2010, 12:59 PM
eumyang
Remember that $\displaystyle 4 = 2^2$, so you could rewrite like this:
\displaystyle \begin{aligned} 3 \cdot 2^x &= 4^{x + 1} \\ 3 \cdot 2^x &= (2^2)^{x + 1} \\ 3 \cdot 2^x &= 2^{2x + 2} \\ \end{aligned}
Can you take it from here?

Regarding the second,
\displaystyle \begin{aligned} 5 \cdot 4^{2x} &= 4 \cdot 2^{6x} \\ 5 \cdot (2^2)^{2x} &= 2^2 \cdot 2^{6x} \\ 5 \cdot 2^{4x} &= 2^{6x + 2} \\ \end{aligned}
Can you take it from here?
• Jul 27th 2010, 01:44 PM
garougt
Quote:

Originally Posted by eumyang
Remember that $\displaystyle 4 = 2^2$, so you could rewrite like this:
\displaystyle \begin{aligned} 3 \cdot 2^x &= 4^{x + 1} \\ 3 \cdot 2^x &= (2^2)^{x + 1} \\ 3 \cdot 2^x &= 2^{2x + 2} \\ \end{aligned}
Can you take it from here?

Regarding the second,
\displaystyle \begin{aligned} 5 \cdot 4^{2x} &= 4 \cdot 2^{6x} \\ 5 \cdot (2^2)^{2x} &= 2^2 \cdot 2^{6x} \\ 5 \cdot 2^{4x} &= 2^{6x + 2} \\ \end{aligned}
Can you take it from here?

Hi, thanks for the reply! However I'm still not sure how to solve for x. I'm suppose to use Logarithmic function, I just don't know what to do with "3" when I use Logarithmic function on 3(2)^x = 4^(x + 1)
• Jul 27th 2010, 01:53 PM
pickslides
$\displaystyle 3 \cdot 2^x &= 2^{2x + 2}$

$\displaystyle 3 &= \frac{2^{2x + 2}}{2^x}$

$\displaystyle 3 &= 2^{2x + 2-x}$

$\displaystyle 3 &= 2^{x + 2}$

$\displaystyle \log_23 &= {x + 2}$

Now can you finish?
• Jul 27th 2010, 04:28 PM
garougt
Quote:

Originally Posted by pickslides
$\displaystyle 3 \cdot 2^x &= 2^{2x + 2}$

$\displaystyle 3 &= \frac{2^{2x + 2}}{2^x}$

$\displaystyle 3 &= 2^{2x + 2-x}$

$\displaystyle 3 &= 2^{x + 2}$

$\displaystyle \log_23 &= {x + 2}$

Now can you finish?

Yes this helped! Well I had to still figure out how it works, but looking at this helped me still.

3*(4)^x also means Log 3 + (x) Log 4. Thanks very much for the replies, very much appreciated =)
• Jul 27th 2010, 05:35 PM
pickslides
Quote:

Originally Posted by garougt
Well I had to still figure out how it works,

You need to remember that $\displaystyle a^b = c\implies b = \log_ac$