Question

Solve log (x+2) + log (x-2) = log 3 + 3 log 4

Answer 1

Hey there!

We know that,

log m + log n = logmn.

and

n logm = log m^n

Now,

$log (x+2) + log (x-2) = log 3 + 3 log 4 \\ \\ log (x+2) (x-2) = log 3 + log 4 ^{3} \\ \\ log (x ^{2 } - 4 ) = log( 3 \times 4 ^{3} ) \\ \\ {x}^{2} - 4 = 192 \\ \\ {x}^{2} = 196 \\ \\ x = ±14$

Therefore , x = 14 or -14 .

Hope helped! :)
Thanks for asking this question, It's #4500 :)

Answer 2

log (x+2) + log(x-2) = log 3 + 3 log 4

We know ,

log mn = log m + log n

Using this,

log (x+2)(x-2) = log 3 + 3 log 4

We know,

m log n = log n^m

Using this in RHS

log (x+2)(x+2) = log 3 + log 4^3

log (x+2)(x-2) = log 3 + log 64

Again using,

log m + log n = log mn

We get,

log (x+2)(x-2)  = log 3×64

log (x+2)(x-2) = log 192

Cancelling log to the base 10 from both sides of the equation , we get :

(x+2)(x-2) = 192

Using identity , (a+b)(a-b) = a^2-b^2 in LHS, we get :

x^2 - 4 = 192

x^2 = 196

x = √196

x = ±14 ( Final Answer )