which approach is the right one..please?
\(2log x - log900=-2\)
and then carry on OR
\(log(x+2)^2= log 900\)
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
My solution is a bit shorter.
2log x+2 =logx2+2
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?
2log3 + 2 = 2*0.477121255 + 2 = 0.954242509 + 2 = 2.954242509
log900 = 2.954242509
x = 3 is correct
your second line is not equivalent to your first line.
Why do you think
can be simplified to
\(log(x+2)^2= log 900 \)
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...
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!