Hi friends...

which approach is the right one..please?

\(2logx+2=log 900\)

\(2log x - log900=-2\)

and then carry on OR

\(2logx+2=log 900\)

\(log(x+2)^2= log 900\)

and then carry on.

I guess my question is this: Is one allowed to "break" the coefficient of a log?. Am I allowed to seperate the "2" from the "x"?

Thank you all once again for the help...

juriemagic Nov 14, 2019

#1**+2 **

Solve for x:

(2 log(x))/log(10) + 2 = log(900)/log(10)

Rewrite the left hand side by combining fractions. (2 log(x))/log(10) + 2 = (2 (log(x) + log(10)))/log(10):

(2 (log(x) + log(10)))/log(10) = log(900)/log(10)

Divide both sides by 2/log(10):

log(x) + log(10) = log(900)/2

Subtract log(10) from both sides:

log(x) = log(900)/2 - log(10)

-log(10) + log(900)/2 = log(1/10) + log(sqrt(900)) = log(1/10) + log(30) = log(30/10) = log(3):

log(x) = log(3)

Cancel logarithms by taking exp of both sides:

**x = 3**

Guest Nov 14, 2019

#2

#3**+2 **

hi guys,

i do appreciate the answers. So, is doing this then wrong?

\(2logx+2=log900\)

\(log(x+2)^2=log900\)

\((x+2)^2=900 \)

\(x+2= \sqrt900\)

\(x+2= \pm30\)

\(x=28\)

or

\(x=-32\) which is not applicable

so \(x=28\)

juriemagic
Nov 14, 2019

#4

#5**+2 **

Isee,

however I did a test..

if i substitute x with 3, which gives us 2 log5 = log900, then things do not balance since 2 log 5 equates to 1,397, while log 900 equates to 2.954. However, with x being 28, the equation becomes 2 log30, which is the same as log 900..this is confusing?

juriemagic
Nov 14, 2019

#6**+2 **

2log3 + 2 = 2*0.477121255 + 2 = 0.954242509 + 2 = **2.954242509**

log900 = **2.954242509**

**x = 3 is correct**

Omi67 Nov 14, 2019

#7**+3 **

Thanks for this...but this confuses me...why is it mathematically incorrect to do the sum the way I did..which gives x to be 28..then add 2??

juriemagic
Nov 14, 2019

#8**+1 **

your second line is not equivalent to your first line.

Why do you think

\(2logx+2=log 900\)

can be simplified to

\(log(x+2)^2= log 900 \)

--------------

\(2logx+2=logx^2+2=2+log(x)^2\)

Melody
Nov 14, 2019

#10**+2 **

Hi Melody,

I always thought that after a log, everything following is part of that log function....I have learnt now that only the immediate value following the log, is applicable to the log function, and all other terms really stand "loose' from that function...Thank you very much for teaching me something again today....I really do appreciate...

juriemagic
Nov 14, 2019

#9**+2 **

It depends on the original format of the question. If it was originally written as 2*log(x+2) then no, you can't treat the two like a normal interger until you get rid of the logarithm. If, however, it was written 2*log(x)+2, then the two is not a part of the logarithm and can be treated as a normal interger because it is one. Once you determine the right format, it's merely a matter of stacking the logs and solving algebraically.

Hope this helped!

-Daewei

Daewei Nov 14, 2019

#11