Answer:
x = -1, x ≈ 0.7017
Step-by-step explanation:
You want the solutions to the equation ...
[tex](0.5)^{x-2}-\left(\dfrac{1}{9}\right)^{\dfrac{3x}{2}}=5^{x+3}-44[/tex]
We can simplify this equation to a form where it is a sum of exponential terms. Subtracting the right side expression, we get the form f(x) = 0.
[tex](2^{-1})^{x-2}-(3^{-2})^{3x/2}=5^{x+3}-44\\\\(2^2)(2^{-x})-3^{-3x}-(5^3)(5^x)+44 = 0\\\\4(2^{-x}) -(3^3)^{-x}-125(5^x)+44 = 0\\\\f(x) = 4\cdot2^{-x}-27^{-x}-125\cdot5^x+44=0[/tex]
There are no algebraic methods for solving exponential equations of this sort. Logarithms are essentially useless for solving equations involving sums of exponential terms. Any solution will be a "guess", or will be found by iteration. Even if a calculator does not find the solution directly, it can aid immensely with the iteration process.
We observe that for large negative or large positive values of x, the value of f(x) will be a large negative number. Each of the exponential terms has a horizontal asymptote of 0, so we can make estimates of points where the exponential terms begin to interact.
For negative x, the 27^(-x) term will dominate. It will have a value of -44 where ...
-27^(-x) = -44
-x·log(27) = log(44)
x = -log(44)/log(27) ≈ -1.15
This suggests we should look at a value of f(x) for x = -1:
f(-1) = 4·2 -27 -125/5 +44 = 8 -27 -25 +44 = 0
This tells us x = -1 is a solution.
For positive x, the 5^x term will dominate. It will have a value of -44 where ...
-125·5^x = -44
5^x = (44/125) = 0.352
x = log(0.352)/log(5) = -0.648
This suggests another solution lies between -0.65 and -1. A few values in this region are ...
f(-0.65) ≈ -2.15390974424
f(-0.70) ≈ -0.06350431277
f(-0.75) ≈ 1.49878567493
f(-0.71) ≈ 0.29197350565 . . . . a solution is in the interval [-.71, -.70]
There are a few methods available for iterating solutions to this sort of equation. One of the fastest is Newton's method. Using that method, the next "guess", x' is found from ...
x' = x -f(x)/f'(x) . . . . . . where f'(x) is the derivative of f(x)
For our f(x), the derivative is ...
f'(x) = -4·ln(2)·2^-x +ln(27)·27^-x -125·ln(5)·5^x
The calculator display in the attachment shows the above function evaluations where Y0(x) = f(x), and it shows several iterations starting with x = -0.7 where Y1(x) = x -f(x)/f'(x), the Newton's method iterator.
The other solution is found to be x = -0.70174358431.
The solutions are x = -1 and x ≈ -0.70174358431.
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Additional comment
The second attachment shows a graph of the function f(x) we defined above. It also displays the x-intercepts (solutions). It's not clear whether this constitutes a calculator solution or not. Even if the calculator does not provide the final answer, the graph gives a good idea of where to begin iterations.
The equation for f(x) can be rewritten to a form useful for iteration:
[tex]g(x)=\log_5(0.008(4\cdot2^{-x}-27^{-x}+44))\qquad\text{converges on -0.7017}\\\\g(x)=-\log_{27}(4\cdot2^{-x}-125\cdot5^x+44)\qquad\text{converges on -1.0}[/tex]
Iteration using these equations is far slower than Newton's method.
