How to Solve an Exponential Equation with Different Bases

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[@Nothingx303]

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

[@alextang4688]

The 1st step, I will take log. 😉😉😉😉😉😉

[@paulortega5317]

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

[@tontonbeber4555]

Apply rule nr 1 of youtube math videos ... solution is always obvious and integer... like ... x=2
Then of course unique by monotony

[@cuborubik2057]

This equation has two solutions -0.43 and 2

[@nitingl4730]

good

[@toveirenestrand3547]

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.

[@scottleung9587]

I got x=2 right away, but the other solution required more work I didn't have time for.

[@nicholasngo5428]

X=2

[@rakenzarnsworld2]

x = 2
2*25=50

[@carly09et]

quick and dirty gets 2 as a solution - then it gets into ln solutions

[@serhiislobodianiuk776]

Left side is increasing, x=2 is the only solution.

[@SyberMath]

Are you sure?

[@serhiislobodianiuk776]

@@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!

[@sollevi9846]

I thought 2 x 25 would be a nice solution so checked x=2 and voila!

[@s1ng23m4n]

Log both sides and we are almost done)

[@giuseppemalaguti435]

x=-log(5)2....nelle semplificazioni mi sono perso una soluzione..x=2