A little help in this equation please.

Question

The equation goes like :
$$2^n = 3n.$$

There are many graphing calculators using which we can solve it but how can we solve it manually without a calculator?

You posted this with the "discrete-mathematics" tag. Does that mean that we're only considering $n$ to be an integer? In that case, you should clarify that. — Maximal Ideal, Apr 21 '17 at 02:19
You can apply an approximation method like the Newton-Raphson method with $f(x)=2^x-3\cdot x$. — callculus42, Apr 21 '17 at 02:26
@SpiralRain 'n' can have any value not specifically an integer. — Nippun Sharma, Apr 21 '17 at 02:30

score 1 · Answer 1 · answered Apr 21 '17 at 02:18

One method which maybe used:

Let $f(x)=2^x-3x$

You can see that $f(3)=-3$ and $f(4)=4$. Since the function is changing it's value from negative to positive, you can say that a solution lies in $(3,4)$.

Also, $f^{'}(x)=2^x \ln2-3 $ which happens to be decreasing till $x$ approximately equal to $2$ and then increases till $\infty$. Now Since this function has no roots in $(0,2)$, You may as well conclude that the only one unique root lie in $x\in(3,4)$ as an increasing funtion won't be cutting the x-axis again.

Hope it helps.

score 0 · Accepted Answer · answered Apr 21 '17 at 07:53

You could check that, if $f(x)=2^x-3x$, there is another root between $0$ and $1$ since $f(0)=1$ and $f(1)=-1$.

For sure, numerical methods should be the way to use. Let us try a single iteration of Newton method $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Using $$x_0=\frac 12\implies x_1=\frac{\log (2)-2}{\log (4)-3 \sqrt{2}}\approx 0.457526$$ Using $$x_0=\frac 72\implies x_1=\frac{4 \sqrt{2} (\log (128)-2)}{8 \sqrt{2} \log (2)-3}\approx 3.33195$$ while the "exact" solutions would be $\approx 0.457822$ and $\approx 3.31318$

A little help in this equation please.

2 Answers2