If $x$ is real find the maximum possible value of $10^x-100^x$ According to the person who gave this question it apparently has something to do with the range of a quadratic expression. But I can't see the connection with a quadratic equation.
So I tried to solve this by finding the maxima of the expression.
But I don't know how to do it as it's an exponential function.
All I can infer from this is that $x$ must be negative.
 A: $$F(x)=10^x-100^x=10^x(1-10^x)$$
Let $$f(a)=a(1-a)$$
$$f'(a)=1-2a$$
the maximum of $ f(a)  $ is $$ f(\frac 12)=\frac 14.$$
Thus, the maximum of $ F(x) $ is $ \frac 14 $ attained for $ x$ such that
$$10^x=\frac 12 = e^{x\ln(10)}$$
A: $10^x - 100^x = 10^x -(10^2)^x = 10^x - (10^x)^2$
Let $y = 10^x$ then $10^x - 100^x = y - y^2$.
To find the maximum of that... well, use calculus.  $f(y) = y-y^2$ so $f'(y) = 1-2y=0 \implies y=\frac 12$ so the maximum value is $y = \frac 12$.
So $y = 10^x = \frac 12$
And $x =\log_{10} \frac 12 = -\log_{10} 2$.
A: The point being made here is that $100^x = (10^x)^2$ - so your expression is really
$$10^x - (10^x)^2$$
or, if we let $y=10^x$ we get
$$y-y^2$$
which is just a quadratic, which you can maximize - though you do have to note that the range of possible $y$ is just positive real numbers, though that's okay because $y-y^2$ is maximized with a value of $1/4$ at $y=1/2$ (i.e. when $x = \log_{10}(1/2)$).
Of course, the usual calculus method works too - if you let $f(x)=10^x-100^x$ and differentiate you get $f'(x)=\ln(10)10^x-\ln(100)100^x$. Setting this to zero and moving things around gives
$$\ln(10)10^x = \ln(100)100^x.$$
You could then solve this by taking a logarithm of both sides, which gives
$$\ln(\ln(10))+\ln(10)x = \ln(\ln(100)) + \ln(100)x$$
which is then a linear equation that solves as
$$x=\frac{\ln(\ln(100))-\ln(\ln(10))}{\ln(10)-\ln(100)}$$
which is equal to $\log_{10}(1/2)$ if you simplify - although hopefully if you get that far in simplifying, you would realize that $100=10^2$ at some point! This method works even when there's no special relationship between the bases.
A: Let $t=10^x$.  Then the expression to be maximised is (for $t>0$): $$t-t^2$$
$$ = \frac14 -(\frac12 - t)^2$$ which clearly has a maximum of $\frac14$
A: @MilesB pointed out that I can turn this into a quadratic by substituting $10^x$  with something else.
Let $t=10^x$.
The given expression becomes,
$$-t^2+t$$
Now we can find the maxima by using a special property of quadratic functions.
A function $f(x)=ax^2+bx+c$ where $a$ is negative is maximum at $x=\frac{-b}{2a}$
So the maximum of $-t^2+t$ is at $t=\frac{1}{2}$. So the maximum possible value of the given expression is
$$\left(\frac{1}{2}\right)^2+\frac{1}{2}$$
$$=\boxed{\frac{1}{4}}$$
A: Alternatively:
$f(x) = 10^x - 100^x$
$f'(x) = \ln 10*10^x - \ln 100* 100^x$.
$f'(x) = 0$ means
$\ln 10*10^x - \ln 100* 100^x=0$ means
$\ln 10*10^x =\ln 100* 100^x$
$\frac {\ln 10}{\ln 100} = \frac {100^x}{10^x}$
LHS is $\log_{100} 10 = \frac 1{\log_{10} 100} = \frac 12$ and RHS is $(\frac {100}{10})^x = 10^x$.
So $10^x =\frac 12$ and $x = \log_{10} \frac 12$.
