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Question Number 197171 by MM42 last updated on 09/Sep/23

2^(log_3 (x^2 +1)) +2×(x^2 +1)^(log_3 2) =12 ⇒x=?

$$\mathrm{2}^{{log}_{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)} +\mathrm{2}×\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{log}_{\mathrm{3}} \mathrm{2}} \:=\mathrm{12} \\ $$$$\Rightarrow{x}=? \\ $$$$ \\ $$

Answered by Frix last updated on 09/Sep/23

a^(log_( b) c) =c^(log_b a) Which is easy to prove: ln (a^(log_b c) )=log_b c ln a =((ln a ln c)/(ln b))= =log_b a ln c =ln (c^(log_b a) ) Here we have 2^(log_3 (x^2 +1)) =(x^2 +1)^((ln 2)/(ln 3)) The equation now is 3(x^2 +1)^((ln 2)/(ln 3)) =12 (x^2 +1)^((ln 2)/(ln 3)) =4 x^2 +1=4^((ln 3)/(ln 2)) =9 x^2 =8 x=±2(√2)

$${a}^{\mathrm{log}_{\:{b}} \:{c}} ={c}^{\mathrm{log}_{{b}} \:{a}} \\ $$$$\mathrm{Which}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{prove}: \\ $$$$\mathrm{ln}\:\left({a}^{\mathrm{log}_{{b}} \:{c}} \right)=\mathrm{log}_{{b}} \:{c}\:\mathrm{ln}\:{a}\:=\frac{\mathrm{ln}\:{a}\:\mathrm{ln}\:{c}}{\mathrm{ln}\:{b}}= \\ $$$$=\mathrm{log}_{{b}} \:{a}\:\mathrm{ln}\:{c}\:=\mathrm{ln}\:\left({c}^{\mathrm{log}_{{b}} \:{a}} \right) \\ $$$$\mathrm{Here}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{2}^{\mathrm{log}_{\mathrm{3}} \:\left({x}^{\mathrm{2}} +\mathrm{1}\right)} =\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}} \\ $$$$\mathrm{The}\:\mathrm{equation}\:\mathrm{now}\:\mathrm{is} \\ $$$$\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}} =\mathrm{12} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}} =\mathrm{4} \\ $$$${x}^{\mathrm{2}} +\mathrm{1}=\mathrm{4}^{\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}} =\mathrm{9} \\ $$$${x}^{\mathrm{2}} =\mathrm{8} \\ $$$${x}=\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$

Commented by MM42 last updated on 10/Sep/23

very good . ⋛

$${very}\:\:{good}\:.\:\:\:\cancel{\lesseqgtr} \\ $$

Commented by Frix last updated on 10/Sep/23

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