Solve Exponential Variable Equations with Logarithms

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The discussion revolves around solving the equation 5^{3x} - 12^x + 2^{\frac{x}{2}} = 5 for x. Participants express difficulty in applying logarithmic rules effectively, suggesting that taking logarithms complicates the problem further. Numerical approximation is proposed as a potential solution method, with one participant estimating x to be approximately 0.386744. The conversation highlights uncertainty about whether the equation can yield an exact solution or if it inherently requires approximation techniques. Overall, the focus is on finding a reliable numerical approach to solve the exponential variable equation.
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Homework Statement


Solve for x
5^{3x}-12^x+2^{\frac{x}{2}}=5


Homework Equations


logarithm rules?
log(a+b)=?


The Attempt at a Solution


Taking the logarithm of both sides only makes things worse from what I can see.
So really, I don't know how to start.
x\approx 0.386744
 
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Mentallic said:
Solve for x
5^{3x}-12^x+2^{\frac{x}{2}}=5

hmm … that's 125x - 12x + (√2)x = 5 …

i can't see any way of solving that except by numerical approximation :redface:
 
Is this possibly due to the lack of understanding with this maths or because it cannot be solved with an exact answer? Such as 'exact' answers with a series of logarithms, even though the logs themselves are approximated.
Any idea how I could go about obtaining a reasonably accurate numerical approximation?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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