Log attack
Solve these equations.
Problem
Solve the equation $a^x + b^x = 1$ where $0< a, b < 1$ and $a
+ b < 1$, in the special cases:
(i) $a = b\quad $ (ii) $a = {1\over 2}, \ b={1\over 4}\quad $
You can find exact solutions to the equation $a^x + b^x = 1$ in special cases like (i) and (ii).
(i) $a = b\quad $ (ii) $a = {1\over 2}, \ b={1\over 4}\quad $
You can find exact solutions to the equation $a^x + b^x = 1$ in special cases like (i) and (ii).
More generally you will need to use a numerical method for
finding approximate solutions. See
Equation Attack.
Student Solutions
Case (i)
If $a = b$ the equation $a^x + b^x = 1$ where $0< a, b < 1$ and $a + b < 1$ becomes $2a^x =1 $. Then $\log 2 + x \log a = 0$ so the solution can be given in three equivalent forms: Note we can use natural logarithms here or logarithms
to any base.
Case (ii)
Solve $a^x + b^x = 1$, where $a=1/2$ and $b=1/4$. The equation is: Here you could substitute $y = (1/2)^x$
but Trevor multiplied by $4^x$ and used the substitution $y = 2^x$.
By Trevor's method: So
and the equation becomes: $y + 1= y^2$ or $y^2- y- 1 = 0$. Using
the quadratic formula, the negative solution to the quadratic can
be ignored as $2^x$ is never negative, so the solution is: (note this equals the golden ratio, $\phi $).
Therefore $2^x = \phi$ and so $x\ln 2 = \ln \phi$ and the solution
is: giving $x = 0.69424$ to 5 significant figures.
If $a = b$ the equation $a^x + b^x = 1$ where $0< a, b < 1$ and $a + b < 1$ becomes $2a^x =1 $. Then $\log 2 + x \log a = 0$ so the solution can be given in three equivalent forms:
Case (ii)
Solve $a^x + b^x = 1$, where $a=1/2$ and $b=1/4$. The equation is: