Solve-for-real-x-

Facebook Tweet Pin

Question Number 204742 by York12 last updated on 26/Feb/24

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:{x} \\ $$

Commented by York12 last updated on 26/Feb/24

Commented by TonyCWX08 last updated on 27/Feb/24

$${x}\approx\mathrm{0}.\mathrm{933361} \\ $$

Commented by Ghisom last updated on 27/Feb/24

using intuition because it′s not exactly solveable if there′s no easy solution both sides 3^(something) rhs would be nice if x had factors 2^p 7^q log_7 ((2^(p/16) 7^(q/16) )/2^(1/4) )=log_7 (2^((p−4)/16) ) +q very nice if p=4 and q=±3 ⇒ ⇒ rhs=(√3) or rhs=((√3)/3) now try x=2^4 7^3 or x=2^4 7^(−3) in lhs ⇒ x=2^4 7^(−3) =((16)/(343))

$$\mathrm{using}\:\mathrm{intuition}\:\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{exactly} \\ $$$$\mathrm{solveable}\:\mathrm{if}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{easy}\:\mathrm{solution} \\ $$$$\mathrm{both}\:\mathrm{sides}\:\mathrm{3}^{\mathrm{something}} \\ $$$$\mathrm{rhs}\:\mathrm{would}\:\mathrm{be}\:\mathrm{nice}\:\mathrm{if}\:{x}\:\mathrm{had}\:\mathrm{factors}\:\mathrm{2}^{{p}} \mathrm{7}^{{q}} \\ $$$$\mathrm{log}_{\mathrm{7}} \:\frac{\mathrm{2}^{{p}/\mathrm{16}} \mathrm{7}^{{q}/\mathrm{16}} }{\mathrm{2}^{\mathrm{1}/\mathrm{4}} }=\mathrm{log}_{\mathrm{7}} \:\left(\mathrm{2}^{\left({p}−\mathrm{4}\right)/\mathrm{16}} \right)\:+{q} \\ $$$$\mathrm{very}\:\mathrm{nice}\:\mathrm{if}\:{p}=\mathrm{4}\:\mathrm{and}\:{q}=\pm\mathrm{3}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{rhs}=\sqrt{\mathrm{3}}\:\mathrm{or}\:\mathrm{rhs}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\mathrm{now}\:\mathrm{try}\:{x}=\mathrm{2}^{\mathrm{4}} \mathrm{7}^{\mathrm{3}} \:\mathrm{or}\:{x}=\mathrm{2}^{\mathrm{4}} \mathrm{7}^{−\mathrm{3}} \:\mathrm{in}\:\mathrm{lhs} \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{2}^{\mathrm{4}} \mathrm{7}^{−\mathrm{3}} =\frac{\mathrm{16}}{\mathrm{343}} \\ $$

Commented by York12 last updated on 27/Feb/24