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Question Number 200877 by mathlove last updated on 25/Nov/23

Commented by mr W last updated on 25/Nov/23

we know a^(log_a  b) =b, so we can “see”  that x=3 is a root.

$${we}\:{know}\:{a}^{\mathrm{log}_{{a}} \:{b}} ={b},\:{so}\:{we}\:{can}\:``{see}'' \\ $$$${that}\:{x}=\mathrm{3}\:{is}\:{a}\:{root}. \\ $$

Commented by mathlove last updated on 25/Nov/23

thanks

$${thanks} \\ $$

Commented by mathlove last updated on 26/Nov/23

how csn solution mr?

$${how}\:{csn}\:{solution}\:{mr}? \\ $$

Commented by mr W last updated on 26/Nov/23

if it is not obvious to you, i.e. if you  can not “see” the solution, then you  can not solve it at all. there is no  general method to solve such an  equation. for example the root from   x^(log_2  3) =(√(x+1)) is not obvious, so you  can not solve it (exactly).

$${if}\:{it}\:{is}\:{not}\:{obvious}\:{to}\:{you},\:{i}.{e}.\:{if}\:{you} \\ $$$${can}\:{not}\:``{see}''\:{the}\:{solution},\:{then}\:{you} \\ $$$${can}\:{not}\:{solve}\:{it}\:{at}\:{all}.\:{there}\:{is}\:{no} \\ $$$${general}\:{method}\:{to}\:{solve}\:{such}\:{an} \\ $$$${equation}.\:{for}\:{example}\:{the}\:{root}\:{from} \\ $$$$\:{x}^{\mathrm{log}_{\mathrm{2}} \:\mathrm{3}} =\sqrt{{x}+\mathrm{1}}\:{is}\:{not}\:{obvious},\:{so}\:{you} \\ $$$${can}\:{not}\:{solve}\:{it}\:\left({exactly}\right). \\ $$

Commented by mathlove last updated on 26/Nov/23

ok thanks

$${ok}\:{thanks} \\ $$

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