Question

solve 2x + lnx = 2

Answer 1

Hey yenny2012, I'm not sure if my solution is correct but I'll give it a try
\[\log_{10} 10 = 1 \]so writing ln x as log x, we get
\[\log_{10} x + \log_{10} (10)2x = \log_{10} (10)2\]bringing the RHS term over to left,
\[\log_{10} x + \log_{10} (10)2x - \log_{10} (10)2=0\]Simplifying, we'll get
\[\log_{10} x + \log_{10} 20x - \log_{10} 20=0\]Then using the formulas of Logarithm which is
\[\log_{a} mn = \log_{a} m + \log_{a} n\] and \[\log_{a} (m/n) = \log_{a} m - \log_{a} n\]we'll get
\[\log_{10} x(20x) - \log_{10} 20=0\]then \[\log_{10} [x(20x)/20] = 0 \]cancelling common terms of 20, we're left with \[\log_{10} [x ^{2} ] =0\]Taking ln both sides(rmb e^(log x) = x or e^(ln x) =x),
\[e^{\log_{10} (x ^{2})} = e^0 \]which then gives us
\[x ^{2} = 1\]and so \[x = \pm \sqrt{1}\]so \[x = \pm1\]

please do point out if there's any mistakes! thanks :)

Answer 2

\[\ln x = \log_{ e} x\]
basically log with base of e
so you cannot say " ln x as log x "

Answer 3

Simply differentiate both sides with respect to x:
2+1/x=0

Answer 4

Mani_jha your solution is not right.
consider 3x-3=0
if you differentiate both side you will get
3=0
which is definitely wrong.

Answer 5

But log10(10)2x does not equal log10(20x) !!!!!!!!!!!!!!!!!!!!!

Answer 6

2x + lnx = 2
lnx + 2x -2 = 0
Let us consider the function f(x) = lnx + 2x -2
Well if you study this function you'd see one point where the graph intercepts with the x-axis when y = 0 x = 1
witch the geometrical interpretation of the solution we'r looking for here,
HOWEVER I do not know any analytical process to solve this kind of equations, though there is this method called "Newton's method"

Answer 7

lnx= 2(1-x)
e^2(1-x)= x

Taking logs with base 10,
2(1-x)log e= log x

Now, log e= 0.434. Substituting, we get,
2*0.434*(1-x)= log x
0.87-0.87x= log x

Now, converting to exponential form,
x= 10^(0.87-0.87x)
But a^(m-n) = a^m/a^n
Therefore, here, we get,
x= (10^0.87)/ (10^0.87x) Cross multiply.
x* (10^0.87x)= 10^0.87
Now, by simple intuition, one can infer that this equation will hold true only if x=1. For any other value of x, x* (10^0.87x) will not equal 10^0.87.

Hence, x=1.

Answer 8

http://www.wolframalpha.com/
You can also see there steps

Answer 9

Beknazar23 I tried really hard to get \[\ln \left( x \right) = \log_{e} x\]
but I could not

Answer 10

salil.s95 I don't agree with the last part of your answer. I try it as soon as I get a TI to see if I can give you a counter example..

Answer 11

\[2x+lnx=2 \]
Remember\[\ln1 = 0\]\[e^{0}=1\]therefore x=1\[2\times1+0=2\]

Answer 12

I graphed it and I found a better solution for it.
we know that the domain is x>0.

case 1:
if 0<X<1 then 0<2x<2
then we know that if
0<x<1 then ln(x)<0.
so we can conclude that:
2x+ln(x) < 2.

case 2:
then we substituent 1 in the function and we get
2x+ln(x)= 2+0 =2

case 3:
now we consider that
x>1 then 2x>2 and ln(x)>0
so 2x +ln (x) > 2


So the only solution is x=1.