The only solution is obviously zero, but OK: \(x\log 2 = x\log 3\), or \( x(\log 3 - \log 2) = 0\). Since \(\log 2 \ne \log 3\), the only solution is \(x=0\). How did that fall out? BTW, this is not the problem posed in the OP.
Sorry, I made a typo. Meant to write ... 2^x = 3^(x+1) One of the x's should be x+1 for my solution to be correct. This is a standard problem for my students when studying Log and Exponential equations. Again, the simple ommision of the constant of 50 in the original problem makes it very easy to solve in closed form.
[HR][/HR][HR][/HR] It should be 10 cause 2+3=5�10 squared = 50 cause if u add the two x's that x squared At least thats how I got an answer really didn't know how he wanted it to be answered if was supposed to be a specific way
And this month's Thread Necromancy Award goes to: MathMaster, for reviving a discussion that died 5 1/2 years ago! Congratulations!
Seriously. And I fall for it too because I don't often check the OP's posting date. Then I saw BenTheMan respond and I was like "wait a minute......."