1. ## Logarithms

I understand how to do problems such as (2)(3^x)=(7)(5^x) but I'm stuck on these problems:

a) (4.6)(1.06^(2x+3))=(5)(3^x)

b) (2.67)(7.38^x)=(9.36^(5x-2))

c) (7)(0.43^(2x))=(9)(6^-x)

Any help would be greatly appreciated. Thanks.

2. Originally Posted by andymac
I understand how to do problems such as (2)(3^x)=(7)(5^x) but I'm stuck on these problems:

a) (4.6)(1.06^(2x+3))=(5)(3^x)
You do them in the same way.
$\displaystyle ln \left ( 4.6 \cdot 1.06^{2x + 3} \right ) = ln \left ( 5 \cdot 3^x \right )$
(or you can take $\displaystyle log_a( )$ of both sides where a is your favorite base.)

$\displaystyle ln(4.6) + (2x + 3) \cdot ln(1.06) = ln(5) + x \cdot ln(3)$
which is just a linear equation for x. Solve for x.

-Dan

3. Sorry, I'm still not getting the right answer when I solve for x. Could you show me the steps you would take in order to do so? Thanks, and sorry about the double post.

4. Originally Posted by topsquark
You do them in the same way.
$\displaystyle ln \left ( 4.6 \cdot 1.06^{2x + 3} \right ) = ln \left ( 5 \cdot 3^x \right )$
(or you can take $\displaystyle log_a( )$ of both sides where a is your favorite base.)

$\displaystyle ln(4.6) + (2x + 3) \cdot ln(1.06) = ln(5) + x \cdot ln(3)$
which is just a linear equation for x. Solve for x.

-Dan
Taking it up from where I left off:
$\displaystyle 2~ln(1.06) \cdot x + 3 \cdot ln(1.06) + ln(4.6) = ln(3) \cdot x + ln(5)$

$\displaystyle (2~ln(1.06) - ln(3))x = ln(5) - 3 \cdot ln(1.06) - ln(4.6)$

$\displaystyle x = \frac{ln(5) - 3 \cdot ln(1.06) - ln(4.6)}{2~ln(1.06) - ln(3)} \approx 0.093094$

-Dan

5. Alright thanks a lot. I get it now.