Anyone good at math? Please help a retard understand how to solve

>>17810319
wrong board post in /sci/ fggt

GL though.

>>17810319
I see you already have so fuck.. bamboozled ay.

>>17810319
symbloab.com
wolframalpha.com

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15KB, 234x250px

>>17810717
symbolab*

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534KB, 2592x1456px

Check out the rules of logarithms

>>17810319
Have you tried reading the material? If you're still too lazy just use symbolab.

all you need to know is
ln(x^b) = b*ln(x)

for example
ln(x^5)=5*ln(x)

>>17810319
it is simple:
you have an equation. left side we will call a right side we will call b (both sides are just numbers).
a = b means that a and b have the exact same value. so whatever operation you apply to BOTH sides will keep the equation true.

now you we have a = a(x) meaning that a is actually a function of something else we call x.
same goes for b = b(x).

luckily both sides use the same function ln(x).
general rule: if you have a function f and want know the x in f(x) you have to find the inverse function f^-1(x) and apply it.

the inverse of ln(x) is e^x. We apply it to both sides:

e^(5*ln(x^2)) = e^(4*ln(x^3) + 6)

right hand side is because of e^a * e^b = e^(a+b) and (a^b)^c = a^(b*c):

e^6 * e^(4*ln(x^3)) = e^6 * e^(ln(x^3))^4 = e^6
(x^3)^4 = e^6 * x^12.

Left hand side is:
e^(5*ln(x^2)) = e^(ln(x^2))^5 = (x^2)^5 = x^10

so we have:
x^10 = e^6 * x^12

<-> x^-2 = e^6
<-> x^2 = 1/(e^6) = e^-6
<-> x = +- e^-6

you are welcome.

>>17810935
><-> x = +- e^-6
me again: the - is not a valid solution because a negative number ^3 is negative and in your initital equation you take the logarithm of it. therefore the negative solution is invalid. (made some small conversion error somewhere)