A Log Equation With Different Bases

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[@SidneiMV]

2nd method. Result in base "e" is always more desired (at least for me)

[@SyberMath]

Thanks for sharing!

[@n0ne0ne]

I love your enthusiasm! Can you make more videos on differential equations?

[@SyberMath]

Absolutely!

[@abhishekchauhan1193]

Please solve log2log2 x=log3log3 x​@@SyberMath

[@honestadministrator]

Writing k for log (3) to the base 2 and z for log (x) to the base 2 we get z / k = 1 + z z = 1/( 1/k - 1) = k /( k -1) = log (3) to the base (3/2)

[@justabunga1]

There is a third method that you can also do, which is using the properties of logarithm on the right hand side, which is log_3(x)=1+log_2(x). Subtract log_2(x) to both sides, which is log_3(x)-log_2(x)=1. Notice that if you use the change of base formula as log(x)/log(3)-log(x)/log(2)=1, we can go ahead and multiply by its LCD, which is log(2)log(3). This means log(2)log(x)-log(3)log(x)=log(2)log(3). Factor out log(x) and using properties of logarithms), which is log(2/3)log(x)=log(2)log(3). Divide both sides by log(2/3), so log(x)=log(2)log(3)/log(2/3), and x=2^(log_(2/3)(3)) or 3^(log_(2/3)(2)).

[@SPV66]

@justabunga1 Wow! This was an interesting 3rd method. Thanks.

[@roger7341]

I fell over trying to follow the third method.

[@MarshallNyotAlek]

Where did you get 5 that you subtracted from 2x

[@CriticSimon]

That’s just an example

[@houssameoubnini265]

Good video ❤

[@SyberMath]

Glad you enjoyed! ❤️

[@rakenzarnsworld2]

x = 1/6

[@SyberMath]

Are you sure?

[@mcwulf25]

I used 2nd method but using log2. Which gives me 2^(log2(3))/(1-log2(3))

[@SyberMath]

you mean base 2?

[@mcwulf25]

@@SyberMath log base 2, yes. Imagine that that 2 is a subscript!

[@scottleung9587]

I got 10^((log2*log3)/(log2-log3)) as the answer - would that count?

[@SyberMath]

sure

[@aswellkalawo]

You are good but too fast

[@giuseppemalaguti435]

x=3^(ln2/ln2/3)=0,1588889...

[@dw8931]

You forgot the parantheses around 2/3... 😉 The correct result is 0.1528818142002...