# [SOLVED] Solve Logaritmic Equation in terms of natural log

• Jul 7th 2009, 12:17 PM
Snowcrash
[SOLVED] Solve Logaritmic Equation in terms of natural log
I need to solve the following equation:

10^(2X+3) = 280

To express in logarithmic, I know I need to do the following:

log(10)^280 = 2X+3

At this point I'm not sure where to go. I can move the three over to the left:

(log(10)^280) - 3 = 2X

Then divide by two:

((log(10)^280) - 3))/2 = X

I get -.276 = X, but I feel like I'm missing a step in determining the answer. Did I do the steps correctly to come to this answer or is something missing?
• Jul 7th 2009, 12:26 PM
Prove It
Quote:

Originally Posted by Snowcrash
I need to solve the following equation:

10^(2X+3) = 280

To express in logarithmic, I know I need to do the following:

log(10)^280 = 2X+3

At this point I'm not sure where to go. I can move the three over to the left:

(log(10)^280) - 3 = 2X

Then divide by two:

((log(10)^280) - 3))/2 = X

I get -.276 = X, but I feel like I'm missing a step in determining the answer. Did I do the steps correctly to come to this answer or is something missing?

$10^{2x + 3} = 280$

$\ln{10^{2x + 3}} = \ln{280}$

$(2x + 3)\ln{10} = \ln{280}$

$2x + 3 = \frac{\ln{280}}{\ln{10}}$

$2x = \frac{\ln{280}}{\ln{10}} - 3$

$x = \frac{1}{2}\left(\frac{\ln{280}}{\ln{10}} - 3\right)$.
• Jul 7th 2009, 12:32 PM
Snowcrash
Great - I see I did not use the natural logarithmic form, but the answer is the same. (.276) Thank you!
• Jul 7th 2009, 01:25 PM
Prove It
Quote:

Originally Posted by Snowcrash
Great - I see I did not use the natural logarithmic form, but the answer is the same. (.276) Thank you!

No problem.

My personal preference is always the natural logarithm, but in this case, your answer might look a bit neater if you use the logarithm with base 10.

$10^{2x + 3} = 280$

$10^{2x + 3} = 28\cdot 10$

$\log_{10}{10^{2x + 3}} = \log_{10}{(28\cdot 10)}$

$(2x + 3)\log_{10}{10} = \log_{10}{28} + \log_{10}{10}$

$2x + 3 = \log_{10}{28} + 1$

$2x = \log_{10}{28} - 2$

$x = \frac{1}{2}\log_{10}{28} - 1$.