Solving A Quick and Easy Logarithmic Equation

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#ChallengingMathProblems #LogarithmicEquations

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[udbhaw]

Alternative method:-
a^log b=b^log a
Here, x^log 25=25^log x=k
given, x^log 25+25^log x=10
k+k=10,
k=5
Now compare 25^log x=5
(5)² ^ log x=5
5^log x²=5
log x²=1
x²=10
x=√10.
Thank you ☺☺

[volsyb]

It’s enough to know this identity a^log b = b^log a. And you can solve it in seconds.

[gniedu]

In 1st method when you have 10^(y*log 25) you could simply make (10^log 25)^y, which turns to 25^y. You would then have 25^y + 25^y = 10, which leads to y=1/2. Just a shortcut. Anyway great video

[snejpu]

Hey, when you have 10^logz + z = 10, you do not need a substitution for t, because 10^logz is just z. So, 2z=10, z=5 straight. Another substitution leads to final result, but it is just like z=t after all. But cool problem, I wasn't trying to solve it. : )

[hedayaty]

Using identity x^(log_b n) = n^(log_b x) makes it really easy!
Basically identity says both terms are the same. They you have two methods, use x in the base or in the exponent.

1st method:
x^log 25 + 25^log x = 25^log x + 25^log x = 2 * 25^log x = 10 => 5^2.log x = 5 => x=sqrt(10)
2nd method:
x^log 25 + 25^log x = x^log 25 + x^log 25 = 2 * x^25 x = 10 => x^log25 = 5 => ... =>x=sqrt(10)

[zygoloid]

It's a standard log property that x^log y = y^log x.

Proof: x^log_b y = (b^log_b x)^log_b y = b^(log_b x * log_b y) = (b^log_b y)^log_b x = y^log_b x

[vishalmishra]

a^log(b) and b^log(a) are equal since their logs are log(b).log(a) = log(a).log(b). Therefore each term is 10/2 = 5. Taking log, we get -
log(5) = log(x)log(25 = 5^2) = 2.log(x).log(5) => log(x) = 1/2 => x = 10^.5 = sqrt(10)

[musicsubicandcebu]

The essence of solving this problem is realizing that the two terms on left side are equal.

[piman]

"I'll be presenting two methods - let's start with the first one." Impeccable logic!

[justabunga]

Another method besides what I saw from here is to use the properties of exponent/logarithm using the base and exponent to switch. The equation turns into 25^log(x)+25^log(x)=10. This means that 2(25^log(x))=10. Dividing both sides by 2 gives you 25^log(x)=5. Turn the left side of the equation as (5^2)^log(x)=5. Since the base are the same, then the exponents are equal to each other. Using properties of exponent, 2log(x)=1. Dividing both sides by 2 is going to be log(x)=1/2. Therefore, x=10^(1/2)=√(10).

[elias]

Let log(x) = y, then x = 10^y.
So (10^y)^log(25) + 25^y = 10
10^[log(25)*y] + 25^y = 10
10^log(25^y) + 25^y = 10
25^y + 25^y = 10
25^y = 5, which means y = 1/2.
But x = 10^y, so x = 10^(1/2) = sqrt(10)
Nice.

[dariosilva]

10^(log(z)) = z. What was that long detour?

[mirkopacchioni]

The first method reminded me of the film inception, a replacement within a replacement within another one, very interesting. Nice video as always.

[Ozymandi_as]

a^log(b) → log(b).log(a) →b^log(a)
Gives the general identity:
a^log(b) = b^log(a)

[getjinxed]

a ^log (b) = b ^log (a), which means 25 log x + 25 log x = 10 25 log x = 5, log x = 1/2 since 25^1/2 = 5, and then we ge t the x = 10 ^1/2

[adandap]

I thought both these methods were more complicated than they needed to be. Write x^(log25) = 10^[log(x^(log25))] = 10^(log25 logx) = 25^(logx) and then finish it off.

[rayaneblii]

U'r amazing with ur multiple methods

[shauryasingh]

Using the property

a^log(b) = b^log(a), you could write the equation as

25^logx + 25^logx = 10

=> 25^logx = 5 = 25^(1/2)
=> logx = 1/2
=> x = root(10)
:D

[karryy]

As everyone said, I solved it in 6 seconds.
It's easy to know that 25^logx=x^log25
So 25^logx=5
<=> logx=1/2
<=> x=(∛1000)^0.5

[philipkudrna]

Two Z or not two z! Hillarious! I like that you comment it as if you were just solving it as you speak instead of choosing the more professor-like approach…!