# logarithims question, any help appreaciated

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• Sep 20th 2009, 12:46 PM
paul90900
logarithims question, any help appreaciated
I am stuck with a logarithims problem that i just can't seem to solve.
The problem is,

Determine x in

10 to power 2x = 5e to power 2x + 2

The two different bases are confusing me, can any one offer any suggestions on how to go about solving this? am i missing something obvious?

Any help would be appreaciated
• Sep 20th 2009, 12:58 PM
skeeter
Quote:

Originally Posted by paul90900
I am stuck with a logarithims problem that i just can't seem to solve.
The problem is,

Determine x in

10 to power 2x = 5e to power 2x + 2

The two different bases are confusing me, can any one offer any suggestions on how to go about solving this? am i missing something obvious?

Any help would be appreaciated

$10^{2x} = 5e^{2x+2}$

natural log of both sides ...

$\ln(10^{2x}) = \ln(5e^{2x+2})$

$2x \cdot \ln(10) = \ln(5) + 2x + 2$

$2x \cdot \ln(10) - 2x = \ln(5) + 2$

$2x[\ln(10) - 1] = \ln(5) + 2$

$x = \frac{\ln(5)+2}{2[\ln(10)-1]}$
• Sep 20th 2009, 02:08 PM
paul90900
It works but i am still a bit confused
Many thanks for the reply but i must admit i'm a bit confused with the line

http://www.mathhelpforum.com/math-he...e0bd017a-1.gif

why is the 2x + 2 added to ln(5)

Sorry if this is a stupid question,
i have only just started up a maths course after 10 years away from education.
• Sep 20th 2009, 02:15 PM
skeeter
$\ln(10^{2x}) = \ln(5e^{2x+2})$

$2x \cdot \ln(10) = \ln(5) + \ln(e^{2x + 2})$

note that $\ln(e^a) = a$
• Sep 20th 2009, 02:20 PM
paul90900
many thanks, i see it now