Given the equation 6 x + 4 x = 9 x 6^x + 4^x = 9^x 6 x + 4 x = 9 x , solve for all x x x .
Solution The idea is to divide the whole equation by 4 x 4^x 4 x :
4 x + 6 x = 9 x ⇔ 1 + 6 x 4 x = 9 x 4 x ⇔ 1 + ( 3 2 ) x = ( ( 3 2 ) 2 ) x = ( ( 3 2 ) x ) 2 \begin{array}{rrl}
& 4^x + 6^x &= 9^x\\
\Leftrightarrow & 1 + \frac{6^x}{4^x} &= \frac{9^x}{4^x}\\
\Leftrightarrow& 1 + \left(\frac{3}{2}\right)^x &= \left(\left(\frac{3}{2}\right)^2\right)^x =
\left(\left(\frac{3}{2}\right)^x\right)^2\\
\end{array} ⇔ ⇔ 4 x + 6 x 1 + 4 x 6 x 1 + ( 2 3 ) x = 9 x = 4 x 9 x = ( ( 2 3 ) 2 ) x = ( ( 2 3 ) x ) 2
In this way we get a quadratic formula in q = ( 3 2 ) x q=\left(\frac{3}{2}\right)^x q = ( 2 3 ) x :
1 + q = q 2 ⇔ q 2 − q − 1 = 0 ⇔ ( q − 1 2 ) 2 = 5 4 ⇔ q = 1 2 ± 5 2 ⇔ q = 1 ± 5 2 \begin{array}{rrl}
& 1 + q &= q^2\\
\Leftrightarrow &q^2 - q - 1 &= 0\\
\Leftrightarrow &(q-\frac{1}{2})^2 &= \frac{5}{4}\\
\Leftrightarrow &q &= \frac{1}{2}\pm\frac{\sqrt{5}}{2}\\
\Leftrightarrow &q &= \frac{1\pm \sqrt{5}}{2}\\
\end{array} ⇔ ⇔ ⇔ ⇔ 1 + q q 2 − q − 1 ( q − 2 1 ) 2 q q = q 2 = 0 = 4 5 = 2 1 ± 2 5 = 2 1 ± 5
Since q = ( 3 2 ) x > 0 q=\left(\frac{3}{2}\right)^x>0 q = ( 2 3 ) x > 0 , we can ignore the negative solution and proceed with the positive
(golden ratio) q = 1 + 5 2 q = \frac{1+ \sqrt{5}}{2} q = 2 1 + 5 . Setting both variants for q q q equal yields
( 3 2 ) x = 1 + 5 2 ⇔ x ⋅ ln ( 3 2 ) = ln 1 + 5 2 ⇔ x = ln 1 + 5 2 ln ( 3 2 ) = 1.1868143 … \begin{array}{rrl}
&\left(\frac{3}{2}\right)^x &= \frac{1+ \sqrt{5}}{2}\\
\Leftrightarrow &x\cdot \ln\left(\frac{3}{2}\right) &= \ln\frac{1+ \sqrt{5}}{2}\\
\Leftrightarrow &x &= \frac{\ln\frac{1+ \sqrt{5}}{2}}{\ln\left(\frac{3}{2}\right)} \\
&&= 1.1868143\dots
\end{array} ⇔ ⇔ ( 2 3 ) x x ⋅ ln ( 2 3 ) x = 2 1 + 5 = ln 2 1 + 5 = l n ( 2 3 ) l n 2 1 + 5 = 1.1868143 …
Solution using Newton Raphson method We can solve this problem also numerically using Newton Raphson method .
Let f ( x ) = 4 x + 6 x − 9 x f(x) = 4^x + 6^x - 9^x f ( x ) = 4 x + 6 x − 9 x , then f ′ ( x ) = 4 x ln 4 + 6 x ln 6 − 9 x ln 9 f'(x) = 4^x \ln 4 + 6^x \ln 6 - 9^x \ln 9 f ′ ( x ) = 4 x ln 4 + 6 x ln 6 − 9 x ln 9 . Since we want the solution
x x x , for which f ( x ) = 0 f(x) = 0 f ( x ) = 0 , we try some small values: f ( 1 ) = 1 f(1) = 1 f ( 1 ) = 1 , f ( 2 ) = − 29 f(2) = -29 f ( 2 ) = − 29 . That means f ( x ) = 0 f(x)=0 f ( x ) = 0 must be
somewhere between 1 and 2, so we start with x 0 = 1.5 x_0=1.5 x 0 = 1.5 :
The next better estimate for x x x is given by x n + 1 = x n − f ( x n ) f ′ ( x n ) x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} x n + 1 = x n − f ′ ( x n ) f ( x n ) :
x n x_n x n f ( x n ) f(x_n) f ( x n ) f ′ ( x n ) f'(x_n) f ′ ( x n ) x n + 1 x_{n+1} x n + 1 1.50000 -4.303062 -21.901330 1.3035 1.30353 -1.105660 -11.560793 1.2079 1.20789 -0.166685 -8.224199 1.1876 1.18762 -0.006124 -7.625605 1.1868 1.18682 -0.000009 -7.602590 1.1868 1.18681 -0.000000 -7.602555 1.1868 1.18681 0.000000 -7.602555 1.1868 1.18681 -0.000000 -7.602555 1.1868