Logarithm Help: solving logartithmic equations

[shaoen01]

Hi,

Can anyone show me the workings a few log equations? I can't seem to be able to get the answer.

Qns: Solve the following equation,

1.
$\displaystyle 6 + \log_{2}x = (\log_{2}x)^2$

My Working:
\(\displaystyle 6 + \log_{2}x = (\log_{2}x)^2

\log_{2}x - (\log_{2}x)^2 = -6\)

Problem: I am stuck at the second line because i don't know how to remove the square root of $\displaystyle (log_{2}x)^2$

2.
$\displaystyle \log_{2}(x+4) - \log_{4}x = 2$

I do not know how to write "ln" symbol and (x + 4)/x in Tex so i am sorry if i did not put my working. I am not sure if i did my working correctly, so if someone can show me how it is done.

Thanks

 

[wantedcriminal03]

Re: Logarithmn Help

I'm no expert but I love logs. If you got answers check to see if I'm correct.

1.
$\displaystyle 6 + \log_{2}x = (\log_{2}x)^2$


$\displaystyle let y = \log_{2}x$

$\displaystyle 6+y=y^2$
$\displaystyle y^2-y-6=0$
(y+2)(y-3)=0
y=3 or -2

$\displaystyle \log_{2}x=3$
$\displaystyle x=2^3 = 8$
$\displaystyle \log_{2}x= -2$
x=2^(-2) = 1/4


Question 2 is too complicated for me to write clearly so I'll let someone else do it :wink:

I think you have to use the change of base rule on log being subtracted so that you can then apply the subtraction rule which means the second log is divided by the first to the same base and work it out from there. I'm not much good at writing about maths in words so I may be saying something completely wrong but.. :roll:

 

[tkhunny]

Use your logarithm rules.

log(a) - log(b) = log(a/b)

Notice that $\displaystyle 4 = 2^{2}$

Show us what you get.

 

[shaoen01]

Hi,

Thanks for your help! I now know how to solve!

 

[stapel]

I now know how to solve!

Complete solution posted here.