I'll Be Proud If You Solve This 

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I find it slightly more intuitive to let a = 5^x and b = 2^x, then you'll get the same factoring situation.

This is one of the basic exercises for grade 12 in Vietnam

I got it! This is pretty much the only one I've ever gotten from your channel this quickly.

damn i hope i could someday solve this types of problems in no time.

I tried to divide them up into prime-factors at first, but I did in the end get x = log_5/2 (2)

really nice, using nearly all of the power tricks i went with logs in earlier steps and got wrong answer or made some mistake somewhere

I think you could improve the final answer a bit more by using logarithms with whole number bases

I legit didn't expect it to be easy You always present us with mind boggling equations!

16^x - 12^x = 9^(x+y) 3^y = √2 x=?

Be proud of me, I got this one. Slightly different approach: a = 5^x and b = 2^x. Moving everything to one side gives a^2 + ab -6b^2 = 0, which factors to (a - 2b)(a + 3b) = 0. Since 5^x > 0 and 2^x > 0 we know a + 3b = 0 has no real solution. Therefore a - 2b = 0 gives the solution. 5^x - 2*2^x = 0 leads to the same answer, or since I have weird ln reflexes mine says the equivalent ln 2 / ln (5/2).

Divide by 4^x solve quadratic x= ln2/ln5-ln2 hope thats right anyway that was a nice question 😉

Awesome video, Brian! What do you think about sharing the manim code of your videos in exchange for a donation on patreon or somewhere else?

Yall saying its challanging meanwhile its the most casual excersise for 11th grade in Armania 😂

I divided both sides by 25^x and let t = (2/5)^x (t>0) and then solved the quadratic equation ( 1 + t = 6t^2 ) then I found t = 1/2 and -1/3 ( which is eliminated because t>0 ), if t = (2/5)^x then (2/5)^x = 1/2 => x = log_2/5 (1/2), that's my way!

25^x + 10^x = 6 x 4^x => (5/2)^(2x) + (5/2)^x - 6 = 0 => (5/2)^x = (-1 +/- √(1+24))/2 = (-1 +/- 5)/2 = 2 or -3. Taking logs on both sides, x = ln(2) / (ln(5) - ln(2))

In India this is very basic for a grade 11th students preparing for jee (joint entrance exam )

I solved it using natural logs (if that's valid).

The answer to this question is 0.4307 to 4 decimal places, as a result of log2/log5=0.4307 that is the correct answer

log _base 5/2_ ?? I mean yes but no one uses that. The natural log is a better choice here. x = ln(2)/(ln(5) - ln(2))

Easy

Here in the first minute of upload

This equation has only one real solution, but it has infinitely many complex number solutions. Here is how to calculate all solutions, and also how to calculate a natural logarithm of a negative real number. (25^x + 10^x)/3 = 2^(2x+1) (5^x * 5^x) + (2^x * 5^x) = 6(2^x * 2^x) 5^x * 5^x 2^x * 5^x -------------- + ------------- = 6 2^x * 2^x 2^x * 2^x ((5/2)^x)^2 + (5/2)^x - 6 = 0 t = (5/2)^x t^2 + t - 6 = 0 t = (-1 +- 5)/2 (t + 3)(t - 2) = 0 Real solutions: (5/2)^x = -3 (no real solutions) (5/2)^x = 2 x = ln(2) / ln(5/2) = ln(2) / (ln(5) - ln(2)) Complex solutions: e^(i*pi) = -1 (Euler's identity) e^0 = 1 C = i*2*pi*n where n E Z e^(i*pi + C) = -1 e^(0 + C) = 1 (5/2)^x = -3 = 3 * e^(i*pi + C) x = ln(3 * e^(i*pi + C)) / ln(5/2) x = (ln(3) + i*pi + C) / ln(5/2) (5/2)^x = 2 = 2 * e^(C) x = ln(2 * e^(C)) / ln(5/2) x = (ln(2) + C) / ln(5/2) Answer: x = (ln(3) + i*pi + i*2*pi*n) / ln(5/2) where n E Z x = (ln(2) + i*2*pi*n) / ln(5/2) if n = 0: x = ln(2) / ln(5/2) (the only real solution)

I got the answer through a different method (completing the square). This one was fun! Here’s my LaTeX in case anybody wants to look, though it’s a bit messy in a comment: 25^x + 10^x &= 3 \cdot 2^{2x + 1} \\\\ 5^{2x} + 2^x 5^x - 3 \cdot 2^{2x + 1} &= 0 \\ (5^x)^2 + 2 \cdot 2^{x-1} \cdot 5^x - 24 \cdot (2^{x-1})^2 &= 0 \\ (5^x + 2^{x-1})^2 &= 5^2 (2^{x-1})^2 \\ 5^x + 2^{x-1} &= 5 \cdot 2^{x-1} \\ 5^x &= 4 \cdot 2^{x-1} \\ 5^x &= 2 \cdot 2^x \\ x &= \log_5 (2) + x \log_5(2) \\\\ \therefore x &= \frac{\log_5(2)}{1-\log_5(2)}

i tried it where did i go wrong ((25^x)+(10^x))=3*(2^(2x+1)) ((25^x)+(10^x))/(2^(2x+1))=3 (2^(2x+1))=((2^2x)*2) subsitute to get (((25^x)/2)+((10^x)/2)))/(2^(2x)=3 take 2xth root of it to get (((25^x)/2)+((10^x)/2)))to the 2xth root/2= 3to the 2xth root *2on both ^2x on both use (a/2)+(b/2)=((a+b)/2) (((25^x)+(10^x))/2)=(2^2x)*3 /3 *2 (((25^x)+(10^x))/3)=(2^(2x+1)) ((25^x)+(10^x))=3*(2^(2x+1)) dang it