# Math Help - Log Questions

1. ## Log Questions

Hi,

I'm needing help with solving two equations using logarithms:

1) 15000=702.712(1.067)^x

I know by "plugging and chugging" into x, the it's somewhere around 48. However, I'd like to know how to solve it the real way. Here's what I'm trying:

Take the log of both sides log 15000 = log 702.712 (1.067) ^ x
Drag the x out front 4.176 = (x) log 702.712 (1.067)

But when I try to solve it any further, I get x equaling around 1 or 2... not right at all.

Am I solving this equation wrong?

Also, there is one more involving Pe^rt (continuous growth formula)

2) 60000 = 10000e^.062t

Because this equation involves e, I take the natural log of both sides:

ln 60000 = ln 10000 e ^ .062 t

I drag .062t out front:

ln 60000 = (.062t) ln 10000

I divide by ln 10000:

1.1945 = .062 t

Divide by .062,

19.266 = t

That's all good, I managed to solve the entire equation, except the teacher specifically told us our answer should come out between t= 25 and t=30. I think the answer should be around t=29

So, I have two equations here. I know what the answer SHOULD be, but it really frustrates me that I can't solve them and come up with the answer myself. Can you guys point out where I might be going wrong?

Thanks,
Ethan

2. Originally Posted by EthanDavis
Hi,

I'm needing help with solving two equations using logarithms:

1) 15000=702.712(1.067)^x

I know by "plugging and chugging" into x, the it's somewhere around 48. However, I'd like to know how to solve it the real way. Here's what I'm trying:

Take the log of both sides log 15000 = log 702.712 (1.067) ^ x
Drag the x out front 4.176 = (x) log 702.712 (1.067)

But when I try to solve it any further, I get x equaling around 1 or 2... not right at all.

Am I solving this equation wrong?
you can't drag the x out like that... first divide both sides by 702.712, then log both sides...

so, $21.34587 = 1.067^x$

$\Rightarrow \log 21.34587 = \log 1.067^x$

$\Rightarrow \log 21.34587 = x \log 1.067$

now continue...

3. Originally Posted by EthanDavis
Also, there is one more involving Pe^rt (continuous growth formula)

2) 60000 = 10000e^.062t

Because this equation involves e, I take the natural log of both sides:

ln 60000 = ln 10000 e ^ .062 t

I drag .062t out front:

ln 60000 = (.062t) ln 10000

I divide by ln 10000:

1.1945 = .062 t

Divide by .062,

19.266 = t
do the same thing i did in the first. as i said, you cannot drag out the power as you did. you have the log of a product, you'd have to separate them before bring down the power. it's much easier to just divide first, so on one side you have just a number to a power, no constant multiplier in front

4. Thank you,

Those were both very silly mistakes. I will remember to divide before pulling the exponent out front.

The answers came out right. Now I'm embarrassed!

Thank you very much!