Solving e^x=ln(x)/log(x)

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#ChallengingMathProblems #ExponentialEquations

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[GirishManjunathMusic]

Before watching the video:
Given:
e↑x = ln(x)/log(x)

Clearly here x cannot be 1 to ensure RHS is defined, so:
log(x)·(e↑x) = ln(x)

Using change-of-base:
ln(x)·(e↑x)/ln(10) = ln(x)

Moving all terms to LHS:
ln(x)·(e↑x)/ln(10) – ln(x) = 0
ln(x)·(e↑x/ln(10) – 1) = 0

But as ln(x) cannot be 0:
(e↑x)/ln(10) – 1 = 0
e↑x/ln(10) = 1
e↑x = ln(10)

Taking natural log on both sides:
x = ln(ln(10)).

[Anmol_Sinha]

One of the few times where the 1st method is easier than the second one

[TechyMage]

swap bases with input
use property(log x/log y = log base y input x)
U will end up with e^x= ln 10
Take natural log once and ur done
x=ln(ln(10))

[konoveldorada]

Hey, hope you are having a nice day :)
So basically, we were taught that
lnx= 2.303 x logx.
So if we substitute ln x in the equation,
We get e^x = 2.303...
By logging
x=ln(2.303)
Which is exactly ln(ln10).
The value is like x= 0.834, a constant
Which satisfies the equation e^x=2.303.

[Skank_and_Gutterboy]

Nice problem! I did method #1. Since e^x is on the left side of the equation, it makes sense to use change-of-base to change everything on the right to natural logs. If you remember that log(x) has an implied base of 10 if none is shown, converting it and simplifying the expression is very easy.

[sabrenhatam]

شكرا عله جهودكم واتمنى لكم مع اقتراب عيد اضحى مبارك ان يسعدكم

[justabunga]

This equation e^x=ln(x)/log(x) can be written using the change of base of formula as e^x=ln(x)ln(10)/ln(x). Notice that ln(x) cancels each other out leaving the equation as e^x=ln(10), which means that x=ln(ln(10)).

[OCEAN-fcwl]

I love this guy’s occasional humor and puns throughout the video. Makes me giggle sometimes

[moeberry]

Good video Syber, in method 2 however when using the substitution you forgot to substitute 10^u for x in e^x. It should look like the following, ln(e^10^u) =ln(ln(10))= 10^u= 10^log(x)=x.

[seanfraser]

lnx/logx = lnx/(lnx/ln10) = ln10

So we have e^x = ln10, thus
x = ln(ln10)

[MichaelRothwell]

Nice video as always.
The graph of ln x/log x can be a bit misleading. It is defined for x>0 where x≠1. It's easy to miss the hole in the graph at x=1.

[chandrashekharmehta]

Bhai maine 10 sec. Me kar diya

Sach me

Yakeen nhi ho rha
Anyways it was easier
😁

[mryip]

e^x = lnx / (lnx / ln 10)
e^x = ln 10
x = ln (ln 10)

[borisfishgal]

I'm watching your videos for a couple of months. You constantly trying to assume that LogX is on base 10. You don't have to assume 🙂
In many countries this has a specific expression: LogX on base 10 is LgX. Again, it's written as LgX - equivalent of LogX on base 10.
LgX 🙂
Thanks

[snejpu]

Hmm, I used the identity twice, but I must have done something wrong. ln(x)/log(x) = log(e). If e^x = log(e), then x = ln(log(e)). But it is incorrect, because x<0. :-( So it's not in the domain...

[ethanthiebaut]

Cool but ur videos would be much better if u wrote on a white page and stopped coloring the equations with such extreme colors.