Explanation:
2.
as a regular function is y = f(x), the inverse function g is simply turning this around x = g(y). it gives us the original x for a calculated y.
but of course, since we have to describe it as a regular function, we need to rename x to y and vice versa to y = g(x). but the principle starts the same.
so, first, we transform the original function into x = g(y). and then we rename the variables.
y = (2x - 1)/5
5y = 2x - 1
5y + 1 = 2x
x = (5y + 1)/2
out then renamed
y = (5x + 1)/2
7.
25 = 5²
125 = 5³
25^(2x+3) = 5^(2×(2x+3))
125^(2x+8) = 5^(3×(2x+8))
5^(2×(2x+3)) = 5^(3×(2x+8))
means simply that
2×(2x+3) = 3×(2x+8)
4x + 6 = 6x + 24
6 = 2x + 24
-18 = 2x
x = -9
8.
log2(10x + 5) - log2(5) = 5
loga(b) - loga(c) = loga(b/c)
so,
log2((10x + 5)/5) = 5
log2(2x + 1) = 5
2x + 1 = 2⁵ = 32
2x = 31
x = 15.5
19.
x + 14/x = -9
multiply both sides by x
x² + 14 = - 9x
x² + 9x + 14 = 0
the general solution for a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 9
c = 14
x = (-9 ± sqrt((9)² - 4×1×14))/(2×1) =
= (-9 ± sqrt(81 - 56))/2 = (-9 ± sqrt(25))/2 =
= (-9 ± 5)/2
x1 = (-9 + 5)/2 = -4/2 = -2
x2 = (-9 - 5)/2 = -14/2 = -7
the 2 solutions are
x = -2
and
x = -7
20.
25 = 5(3x - 4)^(1/3)
the exponent of 1/3 means nothing else than the cubic root.
how to resolve a cubic root ?
by putting everything to the power of 3.
25³ = 5³(3x - 4)
15625 = 125(3x - 4) = 375x - 500
16125 = 375x
x = 43