1. ## Logarithms

Hey guys, I was just wondering how I would solve these 2 problems.

4^x = 10^(x+1)
I wrote this in logarithmic form and got
xlog4=(x+1)log10, but I'm confused what to do since there is an x on both sides.

45e^(.457p) = 95
Don't even know how to approach this one.

2. Originally Posted by Jubbly
4^x = 10^(x+1)
xlog4=(x+1)log10, but I'm confused what to do since there is an x on both sides.
$x \cdot log(4) = x \cdot log(10) + log(10)$

$x \cdot log(4) - x \cdot log(10) = log(10)$

Now factor the LHS

-Dan

3. Originally Posted by Jubbly
45e^(.457p) = 95
$\displaystyle 45 e^{.457p} = 95$

$\displaystyle e^{.457p} = \frac{95}{45}$

Can you take it from here?

-Dan

4. Thanks so much, but I don't know what to do for this and what does LHS stand for?

5. Originally Posted by Jubbly
Thanks so much, but I don't know what to do for this and what does LHS stand for?
LHS ... Left-Hand Side of the equation

$x\log{4} - x\log{10} = \log{10}$

$x(\log{4} - \log{10}) = \log{10}$

finish it.

6. Originally Posted by Jubbly
Thanks so much, but I don't know what to do for this and what does LHS stand for?
LHS means Left Hand Side; take out the x : x[log(4) - log(10)]

You don't know what to do for WHAT?