exponential equation.

[stuart clark]

If \(\displaystyle \lfloor 2^x \rfloor+\lfloor 3^x \rfloor = \lfloor 6^x \rfloor\). Then find value of \(\displaystyle x\)

Where \(\displaystyle \lfloor x \rfloor =\) floor function.

 

[lookagain]

If \(\displaystyle \lfloor 2^x \rfloor+\lfloor 3^x \rfloor = \lfloor 6^x \rfloor\). Then find value of \(\displaystyle x\)

Where \(\displaystyle \lfloor x \rfloor =\) floor function.

You know what would make both sides equal to 0?

Any negative number would make all three terms equal to a value between 0 and 1 at the same time.

So x < 0 is part of the solution.

For x = 0, you get 1 + 1 = 2 on the left-hand side, which is not equal to 1 on the right-hand side.

And for x = 1, the left-hand side is 2 + 3 = 5, but the right-hand side is 6.

Then you would expect there to be real values of x, belonging to 0 < x < 1, where the two sides are equal.

 

[stuart clark]

Thanks lookagain