Question

Log3(1/9) Log2(корень 2/2) 2)inx =3in b + 2in a -3 in c

Answer 1

Log3(1/9)=-2
log2((√2)/2)= -1/2

Answer 2

I assume you want me to simplify the given expression, using the properties of logarithms. Here are the steps to do that:

1. Log3(1/9)

We know that loga(b) = logc(b) / logc(a) for any positive real numbers a, b, and c. Therefore, we can write:

Log3(1/9) = log10(1/9) / log10(3) [using the base conversion formula]

= -log10(9) / log10(3) [because log10(1/x) = -log10(x) for any positive x]

= -2 / log10(3) [because log10(9) = 2]

Therefore, Log3(1/9) can be simplified as -2/log10(3).

2. Log2(корень 2/2)

Here, корень 2 means the square root of 2. So, we can write:

Log2(корень 2/2) = Log2(√2/2) = Log2(√2) - Log2(2)

= 1/2 - 1 [because Log2(√2) = 1/2 and Log2(2) = 1]

= -1/2

Therefore, Log2(корень 2/2) can be simplified as -1/2.

3. 2inx = 3in b + 2in a - 3 in c

We can use the following logarithmic identities to simplify this expression:

a) loga(b) + loga(c) = loga(bc) [product rule] b) loga(b) - loga(c) = loga(b/c) [quotient rule] c) n loga(b) = loga(b^n) [power rule]

Using these identities, we can write:

2inx = 3in b + 2in a - 3 in c

=> inx^2 = in(b^3) + in(a^2) - in(c^3)

=> in(x^2/(b^3*c^3)) = in(a^2)

=> x^2/(b^3*c^3) = a^2 [using the fact that e^in(x) = x for any real x]

=> x = ±a√(b^3*c^3)

Therefore, the given expression can be simplified as x = ±a√(b^3*c^3).