Anonymous at Fri, 15 Nov 2024 01:49:28 UTC No. 16475531

I tried some GRE math subject test practice problems and found out I'm retarded

Anonymous at Fri, 15 Nov 2024 01:57:03 UTC No. 16475537

>>16475531
same desu. was a straight a math minor and I got 10th percentile on my math gre lol

Anonymous at Fri, 15 Nov 2024 02:02:29 UTC No. 16475539

>>16475537
It's brutal. I'm going to keep trying though.
I'm a HS math teacher right now and my students do really well, but I know I'm no genius. I could go on like this but don't want to stagnate.

Anonymous at Fri, 15 Nov 2024 02:16:50 UTC No. 16475545

My online multiple integrals calc 3 test where I put 6^3=192

Anonymous at Fri, 15 Nov 2024 03:06:32 UTC No. 16475594

>>16475519
I've been studying for qualifying exams in my EE dept and getting absolutely shit on by previous probability and information theory exams.

Anonymous at Fri, 15 Nov 2024 03:12:01 UTC No. 16475601

>>16475519
been slacking on precalc in a nutshell desu. Looks like a great little book tho, bless whoever put that in the wiki

Anonymous at Fri, 15 Nov 2024 03:13:14 UTC No. 16475603

Why is log_a(xy) = log_a(x) + log_a(y)?

Consider log_10(10^x):

log_10(10) = 1
log_10(10 * 10) = 2
log_10(10 * 10 * 10) = 3
log_10(10 * 10 * 10 * 10) = 4
log_10(10 * 10 * 10 * 10 *10) = 5

Add two together and you get the multiple.

Back to the first equation, xy = (a^log_a(x) ) (a^log_a(y) ) = a^(log_a(y)+log_a(x))

The definition says if a^z = xy, log_a(xy) = z, so log_a(xy) = og_a(y)+log_a(x)

Anonymous at Fri, 15 Nov 2024 04:16:58 UTC No. 16475673

Define a positive rational exponent as:
[math]x^{a+b/c} = x^a x^{b/c} = x^a\sqrt[c]{x^b}[/math]

Then we consider exponent addition:
[math]x^{a+b/c}x^{d+e/f}[/math]
First add the whole numbers, no need for proof:
[math]=x^{a+d}x^{b/c}x^{e/f}[/math]
Then get the same denominator:
[math]x^{b/c}x^{e/f}=x^{fb/fc}x^{ec/fc}[/math]
The exponents remain equal during the last step:
[math]x^{eb/ec}=({\sqrt[ec]{x}})^{eb} = (({\sqrt[ec]{x}})^{e})^b = ({\sqrt[c]{x}})^{b} [/math]
Rewrite the numerator based on [math](ab)^{1/x}=a^{1/x}b^{1/x}[/math]
[math]=\sqrt[fc]{x^{fb}x^{ec}}[/math]
And perform the final addition:
[math]=\sqrt[fc]{x^{fb+ec}}=x^{(fb+ec)/fc}[/math]
So by splitting the process into more believable steps we understand why exponent addition holds for rational powers.

Anonymous at Fri, 15 Nov 2024 04:22:21 UTC No. 16475677

Define a positive rational exponent as:
[math]x^{a+b/c} = x^a x^{b/c} = x^a\sqrt[c]{x^b}[/math]

Then we consider exponent addition:
[math]x^{a+b/c}x^{d+e/f}[/math]
First add the whole numbers, no need for proof:
[math]=x^{a+d}x^{b/c}x^{e/f}[/math]
Then get the same denominator:
[math]x^{b/c}x^{e/f}=x^{fb/fc}x^{ec/fc}[/math]
The exponents remain equal during the last step:
[math]x^{eb/ec}=({\sqrt[ec]{x}})^{eb} = (({\sqrt[ec]{x}})^{e})^b = ({\sqrt[c]{x}})^{b} [/math]
Rewrite the numerator based on
[math](ab)^{1/x}=a^{1/x}b^{1/x}[/math]
[math]=\sqrt[fc]{x^{fb}x^{ec}}[/math]
And perform the final addition:
[math]=\sqrt[fc]{x^{fb+ec}}=x^{(fb+ec)/fc}[/math]
So by splitting the process into more believable steps we understand why exponent addition holds for rational powers.

Anonymous at Fri, 15 Nov 2024 04:25:13 UTC No. 16475680

Define a positive rational exponent as:
[math]x^{a+b/c} = x^a x^{b/c} = x^a\sqrt[c]{x^b}[/math]

Then we consider exponent addition:
[math]x^{a+b/c}x^{d+e/f}[/math]
First add the whole numbers, no need for proof:
[math]=x^{a+d}x^{b/c}x^{e/f}[/math]
Then get the same denominator:
[math]x^{b/c}x^{e/f}=x^{fb/fc}x^{ec/fc}[/math]
The exponents remain equal during the last step:
[math]x^{eb/ec}=({\sqrt[ec]{x}})^{eb} = (({\sqrt[ec]{x}})^{e})^b = ({\sqrt[c]{x}})^{b} [/math]
Rewrite the numerator based on
[math](ab)^{1/x}=a^{1/x}b^{1/x} [/math]
[eqn]=\sqrt[fc]{x^{fb}x^{ec}}[/eqn]
And perform the final addition:
[eqn]=\sqrt[fc]{x^{fb+ec}}=x^{(fb+ec)/fc}[/eqn]
So by splitting the process into more believable steps we understand why exponent addition holds for rational powers.