Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite

$$\frac{1}{2} \log(x+2)=2$$

I'm decently good at logarithms but this one seems to be tricky, when I did it myself I got a negative decimal as my answer but I'm not 100% confident in it, and I would really appreciate some help!

share|improve this question
2  
Is your equation $(1/2)\log(x+2)=2$ or $1/(2\log(x+2))=2$? – David Mitra Jan 12 at 19:10
@user57055: Welcome to MSE! Please format questions using MathJax as it makes readability much easier. Also, if this is homework, please tag it as such (and it is okay if it is). It is helpful to the MSE community for you to post what you did so they can identify where you may have gone astray. Regards – Amzoti Jan 12 at 19:10
Obtaining a negative value $x$ as a solution is fine, here, for the equation ${1\over 2\log(x+2)}=2$ (where $\log$ is the natural logarithm). Solving this, we obtain the solution $x=e^{1/4}-2$. This is negative; and no doubt you know you can't take the logarithm of a negative number. But we don't do that here: we take the logarithm of $x\color{maroon}{+2}$, which is positive for our $x$. And this solution does "work". It's always a good idea to check your solutions when solving equations. – David Mitra Jan 12 at 19:16
The equation is (1/2)log(x+2)=2 David – user57055 Jan 12 at 19:20
You shouldn't have obtained a negative solution then. (The solution is $e^4-2$ for the natural logarithm, and $10^4-2$ for the common logarithm.) – David Mitra Jan 12 at 19:25
show 6 more comments

1 Answer

up vote 3 down vote

You have

$\frac{1}{2} \log(x+2)=2$

multiply both sides for 2

$\log(x+2)=4$

Now, I suppose the logarithm base is $e$ so, raise $e$ to both sides of the equation

$(x+2)=e^4$

so, $x=e^4-2$.

Similarly, if the base of the logarithm is 10, the answer is $x=10^4-2$

share|improve this answer
Thank you so much, finally put all the pieces together and found out what I did wrong. The log base of 10 is already built into the word log on my TI-84+ so without the base of 10, the problem would be wrong. I had to pull the base of 10 out mechanically and actually put it into the equation. – user57055 Jan 12 at 19:55
@user57055 : Mathematicians usually take $\log x$ to mean $\log_e x$. Some others often take it to mean $\log_{10} x$. – Michael Hardy Jan 12 at 20:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.