How do u solve for x?(if something to the right of the log isn't in parentheses its the base)?

Question

log2(x) / log4(2x) = log8(4x) / log16(8x)

Answer 1

log2(x) / log4(2x) = log8(4x) / log16(8x)

Let log2(x) = a. Therefore, x = 2^a

Let log4(2x) = b. Therefore, 2x = 4^b

or, (2^1)x = (2^2)^b = 2^(2b), so x = 2^(2b - 1)

Let log8(4x) = c. Therefore, 4x = 8^c

or, (2^2)x = (2^3)^c = 2^(3c), so x = 2^(3c - 2)

Let log16(8x) = d. Therefore, 8x = 16^d

or, (2^3)x = (2^4)^d = 2^(4d), so x = 2^(4d - 3)

Equating all the x's, we have :

a = 2b - 1 = 3c - 2 = 4d - 3

From this, in terms of 'a', we find that :

b = (a + 1) / 2

c = (a + 2) / 3

d = (a + 3) / 4

The original equation may be expressed as

a / b = c / d, and by substitution, we obtain :

a / [(a + 1) / 2] = [(a + 2) / 3] / [(a + 3) / 4]

Simplifying sides :

2a / (a + 1) = 4(a + 2) / [3(a + 3)]

Cross-multiplying :

6a(a + 3) = 4(a + 1)(a + 2)

Dividing through by 2, then expanding :

3a^2 + 9a = 2a^2 + 6a + 4

Simplifying :

a^2 + 3a - 4 = 0

Factorising :

(a - 1)(a + 4) = 0

Thus, a = 1 or a = -4

i.e. log2(x) = 1 implies x = 2^1 = 2

or, log2(x) = -4 implies x = 2^(-4) = 1/16

Plugging both x = 2 and x = 1/16 back into the

equation, shows that they are valid solutions.