x-3-x-1-1-4-x-3-x-2-1-3-

Question Number 87637 by john santu last updated on 05/Apr/20

\sqrt[4]{∣ x - 3 ∣^{x + 1}} = \sqrt[3]{∣ x - 3 ∣^{x - 2}}

Answered by TANMAY PANACEA. last updated on 05/Apr/20

∣ x - 3 ∣^{\frac{x + 1}{4}} =∣ x - 3 ∣^{\frac{x - 2}{3}}

\frac{x + 1}{4} = \frac{x - 2}{3}

4 x - 8 = 3 x + 3

x = 11

Commented by john santu last updated on 05/Apr/20

has three solution sir

x = 2; 4; 11

Commented by TANMAY PANACEA. last updated on 05/Apr/20

o k

Commented by Rasheed.Sindhi last updated on 05/Apr/20

W h a t a b o u t x = 3 ? {I s n}^{'} t i t r o o t ?

Commented by john santu last updated on 05/Apr/20

no

Commented by Rasheed.Sindhi last updated on 05/Apr/20

B u t x = 3 s a t i s f i e s t h e g i v e n e q n .

^{4} \sqrt{∣ x - 3 ∣^{x + 1}} =^{3} \sqrt{∣ x - 3 ∣^{x - 2}}

^{4} \sqrt{∣ 3 - 3 ∣^{3 + 1}} =^{3} \sqrt{∣ 3 - 3 ∣^{3 - 2}}

^{4} \sqrt{0^{4}} =^{3} \sqrt{0^{1}}

^{4} \sqrt{0} =^{3} \sqrt{0}

0 = 0