- #1
BvU said:Well, I think you have a problem. Since $$\left (\left ( \sqrt 2 \right )^\sqrt 2 \right )^\sqrt 2 = 2 $$ I think it should stop at $$ 2 = x^{x^x}$$
You are being lured into the reasoning of post #2, but it's a circular reasoning: IF $$2 = 2^{x^{x^{x^{x^{x^{x^{x^{...}}}}}}}} $$ THEN x =2 . If not, then not!
And I was just as dumb as XL...Without parentheses to modify the order of calculation, by convention the order is top-down, not bottom-up:
Note that some computer programs (notably Microsoft Office Excel) associate to the left instead, i.e. a^b^c is evaluated as (a^b)^c.
To solve an exponential equation, you can use logarithms. Take the logarithm of both sides of the equation, then use the properties of logarithms to simplify the equation and solve for the variable.
The general strategy for solving an exponential equation is to isolate the exponential term on one side of the equation, take the logarithm of both sides, use the properties of logarithms to simplify the equation, and then solve for the variable.
Not all exponential equations can be solved algebraically. Some equations may require the use of advanced techniques or numerical methods to find a solution.
Yes, there are special cases when solving exponential equations, such as when the exponential term has a base of 1 or when the exponent is a negative number.
Yes, for example, to solve the equation 2^x = 8, we can take the logarithm of both sides: log(2^x) = log(8). Using the property of logarithms, we can rewrite the equation as x*log(2) = log(8). Then, we can solve for x by dividing both sides by log(2): x = log(8)/log(2). Therefore, x = 3 is the solution to the equation.