# exponential equation with different bases

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?

• 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. – André Nicolas Jan 20 '14 at 17:38
• Is $x$ an integer value or a real value? – SugerBoy Jan 20 '14 at 17:40
• $x$ is a real value – Mark Jan 20 '14 at 17:41

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

• I can't see only why $(4+5)^t=4+5^t$, I come to $3^{2t}=4+5^t$ – Mark Jan 20 '14 at 23:17
• You have $3^{2t}=9^t$. – Macavity Jan 21 '14 at 2:52
• right, thanks a lot I see now – Mark Jan 21 '14 at 12:23

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.

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.