# Thread: logarithims question, any help appreaciated

1. ## 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

2. 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
$\displaystyle 10^{2x} = 5e^{2x+2}$

natural log of both sides ...

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

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

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

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

$\displaystyle x = \frac{\ln(5)+2}{2[\ln(10)-1]}$

3. ## 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

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.

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

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

note that $\displaystyle \ln(e^a) = a$

5. many thanks, i see it now