+1124 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...
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
+1124 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 \(x=28\)
+1124 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?
+12530 2log3 + 2 = 2*0.477121255 + 2 = 0.954242509 + 2 = 2.954242509
log900 = 2.954242509
x = 3 is correct
+1124 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??
+118687 your second line is not equivalent to your first line.
Why do you think
\(2logx+2=log 900\)
can be simplified to
\(log(x+2)^2= log 900 \)
--------------
\(2logx+2=logx^2+2=2+log(x)^2\)
+1124 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...
+27 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
+1124 Hi Daewei,
yes I fully understand this now...thank you for your input...
