Question

log
$log_{x}(5832) = 6$
im stuck at this sum..​

Answer 1

Answer:

x = 3√2 or x = -3√2

Step-by-step explanation:

We must understand that,

logarithms are just another way of writing bases and exponents

For ex:- 3² = 9

Then, in logarithmic equation we write it as

log(3) 9 = 2

Read as "log to the base 3 of 9 = 2"

It means, to what power must the base 3 raised to get 9 = 2

So, Similarly

log(x) 5832 = 6

In exponents we write it as,

x⁶ = 5832

x = ⁶√5832

Now 5832 = 2³ × 3⁶ or 2³ × (-3)⁶

So, ⁶√(2³ × 3⁶) = 3√2

OR

⁶√(2³ × (-3)⁶) = -(3√2)

If you didn't understand that,

Square root to 6 means it has a power of (1/6)

So, x = (2³ × 3⁶)^(⅙)

(a^m × b^n)^c = (a^m)^c × (b^n)^c

= a^(m×c) × b^(n×c)

x = ((2³)^(⅙)) × (3⁶^(⅙))

x = 2^(3×(1/6)) × 3^(6×(1/6))

x = 2^(1/2) × 3¹

x = 3 × √2

Thus,

x = 3√2

Similarly,

x = -(3√2)

x = -3√2

Hope it helped and you understood it........All the best