JProgress wrote:ln x/2 = (ln x)/2. How could this be solved???
x does equal 4. The above posters are correct, but they are missing a solution (which
luckily does NOT work).
ln(x/2)=(lnx)/2
/* multiply both sides by 2 */
2ln(x/2)=lnx
/* manipulate left site of equation using the properties of logarithms power rule */
ln((x/2)^2)=lnx
/* simply */
ln(x^2/4)=lnx
/* raise both sides to the power of e. This gets rid of both the natural logs. */
x^2/4=x
/* multiply both sides by 4 */
x^2=4x
/* subtract 4x */
x^2-4x=0
/* pull out an x */
x(x-4)=0
thus, we get x=4
or x=0.
Let's plug back in to check our solutions... starting with x=4
ln(4/2)=ln(4)/2
/* simplify the left side of the equation and rewrite ln(4) as ln(2^2) */
ln(2)=ln(2^2)/2
/* using the power rule we can rewrite ln(2^2) as 2ln(2) */
ln(2) = 2ln(2)/2
/* simplify to right side of the equation (2 * 1/2 = 1)
we are left with ln(2)=ln(2), so x=4 is
correct!
Now onto x=0
ln(0/2)=ln(0)/2
We can already stop because ln(0) DNE... When x=0 the equation DNE. This is because ln(0)=x can be rewritten as e^x=0, and no value of x satisfies that equation.