solving a logarithmic equation with different bases

How do we solve a log equation with different bases? Here we will see how we can use the change of base formula for logarithm to solve log_4(x)+log_2(x)=6.

[raphberry]

Could you just have done:
ln(x) + 2 ln(x) = 12 ln(2)
3 ln(x) = 12 ln(2)
ln(x) = 4 ln(2) = ln(2^4) = ln(16)
x = 16

[perlnut]

Why not recognize 4 is 2^2 and change log4(x) to log2(x)/log2(4) and continue from there? Much simpler.

[animalfarm]

Just a thought - If you use a base change to "2" instead of "e", you rapidly get to (log2(x))/(log2(4)) + log2(x)=6; 3/2*log2(x)=6 or log2(x)=4, hence x=2^4

[donati]

Why not :
1/2log2X+log2X=6
3log2X=12
log2X=4
X=2^4
X=16

[lostwizard]

As others have noted, observing that the bases are both powers of 2 suggests that the best base to change to would be 2. That then means the denominators in the change of base become nice integers and the result then trivially simplifies to log2(x) = 4. Of course, your way works more generically and shows off more log properties.

[AndreasChristianto]

i prefer:

ln x + 2 ln x = 12 ln 2
3 ln x = 12 ln 2
ln x = 4 ln 2
ln (x) = ln (2^4)
x = 2^4
x = 16

[flamingpaper]

You can do that last step a little easier. When you have
ln(x) + 2ln(x) = 12ln(2)
You can simply add the ln(x) on the left since ln(x) acts like a variable and get 3ln(x) = 12ln(2)
Then divide both sides by 3 to get
ln(x) = 4ln(2)
Then put the 4 into the ln(2) to get
ln(x) = ln(2^4)
Then x = 2^4, which is obviously 16

[kujmous]

I like this video especially because I can see the answer beforehand. I did it differently, but it uses the same reasoning.
LOG4(x) + LOG2(x) = 6
LOG4(x) + 2(LOG4(x)) = 6
3LOG4(x) = 6
LOG4(x) = 2
x = 4² = 16

[yashjakhmola]

Not bragging, but this was too easy. I literally solved it in my head just looking at the thumbnail.

Your videos are awesome tho.

[not_vinkami]

Maybe it's time for us to make it *COMPLEX*

[suniltshegaonkar]

Also that we can convert Log X_4 = 1/2 of Log X_2. So both terms are at Base 2, taking common the Log portion).
3/2* LogX_2 = 6 >>> X^3/2 = 2^6 >>> X= 16.

[mathsplus]

I really enjoyed how you showed how easy to change the base. Great!

[denismilic]

I rarely can solve math problems on this channel in my head, or even solve without pain but this problem is super simple or I got some wierd inspiration. Answer just poped in my head Sheldon style.

log4(x)+log2(x)=6
log2(sqrt(x))+log2(x)=2 +4
log2(sqrt(x)) = 2 and log2(x)=4
x= 2^4 = 16

[kidtherookie]

I found another way!
Note: 2*log_4(x)=log_2(x)
log_2(x)+2*log_2(x)=12
log_2(x^3)=12
x=2^(12/3)=16

[chyawanprash]

I did it like this.
Let y=log4 (x)
Let z=log2 (x)
y+z=6
4^y=x
2^2y=x

2^z=x

2^2y=2^z
2y=z

y+2y=6
3y=6
y=2
log4 x = 2
x=4²
x=16

[millicentatieno]

surely i had a problem in solving that, thanks dude

[tbg-brawlstars]

BlackPenRedPen : Takes 6 mins to solve the world's easiest log question
Me : Solves it in literally 10 seconds in head

[NeedBetterLoginName]

Around 3:30, you could just do ln(x) + 2ln(x) = 3 ln(x) and go from there, quicker and simplier

[masterclash]

A “hidden” rule I found with logs is that if you have log_b(a), it will be equal to log_(b^2)(a^2) or alternatively 2log_(b^2)(a).

This is because if you use your change in base formula and exponentiate the arguments to the same number, then you can use the power rule to take the exponents out of the argument and cancel, of which would leave you with another fraction to apply the change in base formula.

Example:
log_9(4)

ln(4)
———
ln(9)

ln(2^2)
———
ln(3^2)

2ln(2)
———
2ln(3)
Simplify to
ln(2)
———
ln(3)
And
log_3(2)
Hence, log_9(4) = log_3(2)

[mjones]

I think this is easier to use the change of base formula on just the first addend only using log(base 2). This gives
log(base 2)x / 2 + log(base 2)x = 6 → (3/2) log(base 2)x = 6 → log(2)x = 4 → x = 2⁴ = 16.