How to Solve an Exponential Equation with Different Bases
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- Опубликовано: 10 сен 2024
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2nd method is the best but I will prefer 1st if the Exponential Equation is in the form of 4^x+3 = 5^x+5
The 1st step, I will take log. 😉😉😉😉😉😉
Easier to let c = the log 5 of 2. Then x^2 +x(c-2) - 2c = 0 and (x+c)(x-2) = 0; x = 2 or x = -c
Apply rule nr 1 of youtube math videos ... solution is always obvious and integer... like ... x=2
Then of course unique by monotony
This equation has two solutions -0.43 and 2
good
4^[(x - 1)/x]*5^x = 2^[2(x - 1)/x]*5^x = 50 = 2*5^2. Hence 2(x - 1)/x = 1 and also x = 2, which for the equation 2(x - 1) = x yields 2x - 2 = x, so x = 2.
I got x=2 right away, but the other solution required more work I didn't have time for.
X=2
x = 2
2*25=50
quick and dirty gets 2 as a solution - then it gets into ln solutions
Left side is increasing, x=2 is the only solution.
Are you sure?
@@SyberMath Yes, I was wrong. There are two intervals where it is increasing separately: x > 0 and x < 0, and there is another root -log_5(2).
Thank you!
I thought 2 x 25 would be a nice solution so checked x=2 and voila!
Log both sides and we are almost done)
x=-log(5)2....nelle semplificazioni mi sono perso una soluzione..x=2