log₃(x+5)+log₃(x−7)=2, my answer is in the description please check it out?

log₃(x + 5) + log₃(x − 7) = 2
log₃[(x + 5)(x − 7)] = 2
(x + 5)(x − 7) = 3²
x² - 2x - 35 = 9
x² - 2x - 44 = 0
x = {-(-2) ± √[(-2)² - 4(0)(-44)]} / [2(1)]
x = (2 ± √180)/2
x = 1 + 3√5  or  x = 1 - 3√5 (rejected for x - 7 < 0)
Hence, x = 1 + 3√5 ≈ 7.7

Your answer is correct.

log₃(x + 5) + log₃(x - 7) = 2

so, log₃[(x + 5)(x - 7)] = 2  

Hence, (x + 5)(x - 7) = 3²

i.e. x² - 2x - 44 = 0

Using the quadratic formula we get:

x = 1 ± 3√5  

Now, x = 1 + 3√5 is the only valid solution as x + 5 > 0 and x - 7 > 0

so, x = 7.7...to 1 d.p.

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log₃(x + 5) + log₃(x - 7) =2

Use product rule:

log₃[(x + 5)(x-7)] = 2

Exponentiate using 3 on both sides:

(x + 5)(x - 7) = 3^2

(x + 5)(x - 7) = 9

x^2 - 2x - 35 = 9

x^2 - 2x - 44 = 0

Using quadratic formula, the roots are 7.708203932 and -5.708203932.

But, the negative answer is extraneous and thrown out since negative logs are not allowed.

About 7.7, your answer is correct.

Notation: Put log,base3 = L.;
L(x+5) + L(x-7) = 2, ie., L((x+5)(x-7)) = 2, ie., 3^L((x+5)(x-7)) = (x+5)(x-7) = 3^2.
(x+5)(x-7) = 9, ie., x^2 -2x -44 = 0. Then 2x = 2(+/-)D, where D^2 = 4 + 4*44 =;
4*45= 4*9*5 = (6^2)(rt5)^2 = (6rt5)^2. Then x = 1 + 3rt5, rt = ''square root of''.;
Negative root 1-3rt5 is invalid for x since both (x+5) & (x-7) must be positive.;
x = 7.708203933 = 7.7 rounded off to 1 decimal place.