Solve the equation
In10^x=Log e^x +2
Now I got
xIn10 = xLog e +2
x 2.302585093 = x 0.434294481 + 2
x = 0.434294481 +2/2.302585093
so x = 1.0572066
But imputting this into the original equation is incorrect. Where did I go wrong? Help would be much appreciated thank you.
Simon
Do you not understand why you do those things? Your objective is to solve for x- that is, get x alone on one side. Since, in the second case, the only reason x is NOT alone is because it is multiplied by 6, it makes sense to divide by 6 to get 6x/6= 12/6. But that is NOT the case in "6x= 4x+ 2". You could divide both sides by 6- you would get x= (4/6)x+ (2/6)= (2/3)x+ (1/3). But because you still have x on both sides of the equation, that does not help a lot- and it has those annoying fractions. Instead, it is better to get rid of that "4x" on the right side. And because it is added you get rid of 4x by subtracting it from both sides:
6x- 4x= 4x+ 2- 4x or 2x= 2. Now what do you do to find x?
Also, going back to the original problem, you really need to distinguish between "I" (large letter "i") and "l" (small letter "L"). The symbol for the natural logarithm is "ln", not "In".