# Math Help - Solving equations involving logs

1. ## Solving equations involving logs

Help with these two would be grealy appreciated.

2. Originally Posted by Fnus
Help with these two would be grealy appreciated.
2. From equation (2): $2x = 9 - 3y \Rightarrow x = \frac{9-3y}{2}$.

Substitute this into equation (1):

$2^{\frac{9-3y}{2}} \cdot 2^y = 16 \Rightarrow 2^{\frac{9-3y}{2} +y} = 2^4 \Rightarrow \frac{9-3y}{2} +y = 4$ .......

3. Originally Posted by Fnus
Help with these two would be grealy appreciated.
1. On the left hand side, you're missing an argument for the log. Nevertheless:

Note that $2 = \log_5 25$ and so the right hand side is $\log_5 25 + \log_5 2 = \log_5 50$.

The left hand side can also be expressed as a single log. Do it.

Now equate the arguments of the logs on each side and solve for x.

4. Yay, I get y = 1, and x = 3, and I'm pretty sure that checks out.
Thanks a ton. (:

5. Oh, yeah, on the left side I'm missing an x.

But I don't get what it is you're trying to do by $
\log_5 25 + \log_5 2 = \log_5 50
$

6. Originally Posted by Fnus
Oh, yeah, on the left side I'm missing an x.

But I don't get what it is you're trying to do by $
\log_5 25 + \log_5 2 = \log_5 50
$
Then the left hand side can be written $\log_5 [(x-5) x]$.

The right hand side can be written $\log_5 50$.

Therefore $(x - 5) x = 50 \Rightarrow x^2 - 5x - 50 = 0 \Rightarrow x = .....$

Warning: Although the quadratic equation has two solutions, only one is valid (why?).

7. Oh *okay*.

And I get x = -5 or 10, but it can't be -5 because you can't take the logarith of a negative number, I think..

8. Originally Posted by Fnus
Oh *okay*.

And I get x = -5 or 10, but it can't be -5 because you can't take the logarith of a negative number, I think..

9. ## Another way for the 2nd one

$2^x \cdot 2^y=16 \iff 2^{x+y}=2^4$

Now we note from here the bases are the same so the exponents must be equal. So we get $x+y=4$ now we multiply it by -2 and add to the one below to get

$2x+3y=9$
$-2x-2y=-8$
$y=1$

Subbing back in we get
$x=3$