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We have $3^x-5^\frac{x}{2}=4$ My question is what we can do here ? Can we solved it algebraically or we need to notice that $x=2$ and then show that for $x \neq 2$ there aren't any other solutions?

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  • $\begingroup$ There are exceptions, of course, but in general the situation is as you describe. One has to use inspection, or a numerical approximation method, to find a root, and show that there are no others. I am more or less sure that this is the case here. $\endgroup$ – André Nicolas Jan 20 '14 at 17:38
  • $\begingroup$ Is $x$ an integer value or a real value? $\endgroup$ – SugerBoy Jan 20 '14 at 17:40
  • $\begingroup$ $x$ is a real value $\endgroup$ – Mark Jan 20 '14 at 17:41
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Your second approach, viz. showing there are no other solutions would also require some algebra. With $2t = x$, you can write the equation as $$(4+5)^t = 4 + 5^t$$

This is obvious for $t=1$ i.e $x=2$.

For $t > 1$, we have $(4+5)^t > 4^t+5^t > 4+5^t$

for $0 < t < 1$, let $y = \frac1t > 1$, then $(4+5^t)^y > 4^y+5 > 4+5 = \left((4+5)^t\right)^y$

and for $t \le 0$, $(4+5)^t \le 1 < 4 < 4+5^t $

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  • $\begingroup$ I can't see only why $(4+5)^t=4+5^t$, I come to $3^{2t}=4+5^t$ $\endgroup$ – Mark Jan 20 '14 at 23:17
  • $\begingroup$ You have $3^{2t}=9^t$. $\endgroup$ – Macavity Jan 21 '14 at 2:52
  • $\begingroup$ right, thanks a lot I see now $\endgroup$ – Mark Jan 21 '14 at 12:23
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The given equation holds for the roots of the function

$$f(x) = 3^x-\sqrt{5}^x-4$$

The derivate is

$$f '(x)=3^xln{3}-\sqrt{5}^xln{(\sqrt{5}})$$

So, f '(x) > 0 for all x > 0.

So, f(x) has at most one root.

This root can only be calculated with numerical methods, but in this case, it is easy to guess the solution.

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To go one step further, let me consider the equation $3^x-5^\frac{x}{2}=a$ where $a$ is a positive constant.

What Peter showed is that, since the derivative is almays positive, the equation has only one root. In tour case, the problem was simple since, for $a=4$, there is an obvious solution at $x=2$.

Where the problem starts to be different is when there is not obvious solution; except using inspection to try to find two values of $x$ which bracket the solution, the only other solution is to plot the fuction for the same goal. This wille give you a rough estimate I shall write $x_\text{old}$; this value will be the starting point of a root-finder method. Here, for simplicity, I shall use Newton method.

So, for an example, let me chose $a=12345.6789$. Plotting the function shows that the solution is somwhere between $8$ and $9$. So, being lazy, I shall start the iterations at $x_\text{old} = 8$.

As you know, Newton iteration scheme write

$x_\text{new} = x_\text{old} - f(x_\text{old}) / f'(x_\text{old})$

and, after each iteration, $x_\text{old}$ is updated by $x_\text{new}$.

So, for the case I selected, the following iterates will appear : $8.00000$, $8.95595$, $8.69676$, $8.65122$, $8.65003$ and this is the end of the process for six significant figures. For sure, you can continue iterating until you reach the desired level of accuracy.

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