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4306270 No.4306270 [Reply] [Original]

25^x = 5 * 625^x

how. the. fuck. does. x = -0.5

mfw i cant do this. googled but couldnt find the HOW TO part of this...

>> No.4306274

GUESS I R TOOOO SMRT FOR DUM /SCI/ HUR HUR HUR

>> No.4306276
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4306276

25^x = 5 * 625^x

5^2x = 5 * 25^2x

5^2x = 5 * 5^4x

5^ 2x = 5 ^ 4x+1

4x + 1 = 2x

2x = -1

x = -1/2

begone with you.

>> No.4306277

25 is actually 5^2. That should help. You also might see something similar with 625.

By the way, OpenStudy is better for answering questions about babytier mathematics.

>> No.4306279
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4306279

25^x = 5 * 625^x
(5^2)^x = 5 * (5^4)^x
(5^2)^x = 5 * (5^4)^x
5^(2x) = 5 * 5^(4x)
5^(2x) = 5^(4x + 1)

2x == 4x + 1
x = -1/2

sage mofo

>> No.4306281

>>4306279
>>4306276

>Not applying log base 5 to both sides and solve in 1 line
>2012

ishiggy diggy

>> No.4306286
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4306286

>>4306281
>implying that requires any less work

>> No.4306371

>>4306281
>using log when you don't even need to
seriously?

>> No.4306383

>>4306371

>implying the simplicty of the problem doesnt justify the usage of modern algorithms.

>> No.4306388

>>4306383
>2012
>semisubtle trolling
Meh...

>> No.4306391

>>4306371
From
>5^ 2x = 5 ^ 4x+1
to
>4x + 1 = 2x
What do you think is used to guarantee the direct implication?
[megaspoiler]a logarithm[/megaspoiler]

>> No.4306393

>>4306371
You'd have a point if logarithms weren't actually EASIER THAN WHAT YOU'RE DOING, A WELL AS QUICKER!

>> No.4306400

>>4306391
bijectivity of
F: R->R+*
x->5^x

>> No.4306449

>>4306400
Bijectivity is equivalent to the existence of an inverse function. What you're saying here is exactly the same as using log on both sides. In one case, you say apply "log_5" on both sides and use log's properties, in the other, you apply "the inverse function of x->5^x" on both sides using properties of inverse functions. You're just rewording it.

>> No.4306451

>>4306449
there is a difference between construction of something and the existence in itself.