An Exponential Log Equation #algebra

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[robinswamidasan]

Great videos. I find your use of graphs at the end most interesting & useful.

[Viki]

You can write x^(1/lnx) as (e^(lnx))^(1/lnx) and you can multiply the powers to get e out of that entire expression for x>0 and x≠1, that's how I found that out, pretty interesting fact

[spelunkerd]

I've come to the conclusion you choose to present the more interesting and less obvious solution first. This keeps us entertained even when the solution is obvious. In this case, I simply remembered that x^(1/lnx) is e. This led to the simple equation 1=x-1.

[carlb_]

x^(1/ln(x) is always equal to e because when we take the natural logarithm on that expression its equal to 1(except for undefined ln(x)). That is why we can simply write e=e^(x-1). The only way for e=e^(x-1) is for x-1=1 and the only way that works is by x=2.

[SidneiMV]

1/lnx = logₓe

x^logₓe = e

x - 1 = 1 => *x = 2*

[prollysine]

ln(x^(1/lnx))=ln(e^(x-1)), -->, (1/lnx)lnx=x-1, 1=x-1, x=2, test, 2^(1/ln2)=e, e^(2-1)=e, same, OK,