Solve the exponential equation algebraically?

**tsmith** · October 12th 2009, 10:48 AM

Solve the exponential equation algebraically: -14 + 3e^x = 11

So I think I move the -14 to the other side, so now I'm left with
3e^x = 25
What do i do next?

**e^(i*pi)** · October 12th 2009, 10:50 AM

Originally Posted by tsmith

Solve the exponential equation algebraically: -14 + 3e^x = 11

So I think I move the -14 to the other side, so now I'm left with
3e^x = 25
What do i do next?

Divide both sides by 3
Take the natural log of both sides. Do not use a decimal approximation

**tsmith** · October 12th 2009, 10:55 AM

Ok, so
e^x = 25/3
How do I find the natural log of both sides?

**e^(i*pi)** · October 12th 2009, 11:02 AM

Originally Posted by tsmith

Ok, so
e^x = 25/3
How do I find the natural log of both sides?

$ln(x) = log_e(x)$

You don't need to unless you're making a decimal approximation. It is fine to say that $x = ln \left(\frac{25}{3}\right) = 2\ln\left(\frac{5}{3}\right) = 2ln5 - ln3$

Note that I used the laws of logs to simplify. If you're teacher wants you to make a decimal approximation there should be a button marked ln on your calculator

**tsmith** · October 12th 2009, 11:17 AM

My teacher wants me to round the result to 3 decimal places. What do I type in the calculator?

Thank you for your help, by the way

**e^(i*pi)** · October 12th 2009, 11:24 AM

$2*ln(5) - ln(3)$

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