Question - Solving Exponential Equation $2^x + 3^x = 4^x$
Solution:
The image you've provided shows an equation: $$ 2^x + 3^x = 4^x $$.To solve this equation for $$ x $$, we can try to look for integer solutions by inspection or systematically, assuming $$ x $$ could be a real number.First, let's examine the behavior of each term as $$ x $$ increases:- $$ 2^x $$ and $$ 3^x $$ both grow at an exponential rate, but $$ 3^x $$ grows more quickly than $$ 2^x $$ because it has a larger base.- $$ 4^x $$ also grows at an exponential rate, even more quickly than both $$ 2^x $$ and $$ 3^x $$, since $$ 4 = 2^2 $$.Given the nature of the equation, we can see that as $$ x $$ becomes large, $$ 4^x $$ will dominate the left side of the equation making $$ 2^x + 3^x $$ less than $$ 4^x $$.Now, let's consider small values of $$ x $$:- When $$ x = 0 $$, $$ 2^x + 3^x = 4^x $$ because each term equals 1 (any nonzero number to the zeroth power is 1).- When $$ x = 1 $$, $$ 2^x + 3^x = 4^x $$ because $$ 2 + 3 = 5 $$ which is not equal to $$ 4 $$.- When $$ x = 2 $$, $$ 2^x + 3^x = 4^x $$ because $$ 4 + 9 = 13 $$ which is not equal to $$ 16 $$.The case when $$ x = 0 $$ is the only integer solution that satisfies the equation since for $$ x > 0 $$, the left side of the equation will always be smaller than the right side. Let's try to confirm this by solving the equation algebraically.If we did not find any solutions through inspection or want to consider non-integer real number solutions, we'd have to solve the equation algebraically. However, because this equation is transcendental (involves variable exponents across terms that cannot be easily equated), finding such solutions usually requires numerical methods or advanced mathematical techniques beyond primary algebra.In typical pre-calculus or calculus classes, if this problem was given, the teacher would expect the students to find integer solutions through inspection, as described. For non-integer solutions, students would generally be instructed to use a graphing calculator or numerical methods to approximate solutions, but algebraic methods for finding these approximate solutions without a calculator are largely impractical for this form of the equation.As concluded from inspection, the only solution that can easily be found is $$ x = 0 $$.
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