Solve the equation for exact answer. ${9}^{3x+5}=18$

Luz Stokes
2022-07-23
Answered

Solve the equation for exact answer. ${9}^{3x+5}=18$

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${5}^{2x+1}+{5}^{x}-4=0.$

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${2}^{2}+\frac{{3}^{2}}{2}!+\frac{{4}^{2}}{3}!+\dots$ to infinity

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What is the difference between logarithmic decay vs exponential decay?

I am a little unclear on whether they are distinctly different or whether this is a 'square is a rectangle, but rectangle is not necessarily a square' type of relationship.

I am a little unclear on whether they are distinctly different or whether this is a 'square is a rectangle, but rectangle is not necessarily a square' type of relationship.

asked 2022-04-30

Solve for x in the equation $${\mathrm{log}}_{4}\ufeff(1-2x)=4$$.

asked 2022-03-12

What is the simple way to show that

$\frac{N\mathrm{log}\left\{N\right\}}{k\mathrm{log}\left\{k\right\}}\approx {\mathrm{log}}_{k!}\{N!\}$

I tried to use the factorial and the log rules but..

Thanks.

I tried to use the factorial and the log rules but..

Thanks.

asked 2022-05-19

Logarithm Problem : Find the number of real solutions of the equation

$2{\mathrm{log}}_{2}{\mathrm{log}}_{2}x+{\mathrm{log}}_{\frac{1}{2}}{\mathrm{log}}_{2}(2\sqrt{2}x)=1$

My approach :

Solution : Here right hand side is constant term so convert it into log of same base as L.H.S. therefore, 1 can be written as ${\mathrm{log}}_{2}{\mathrm{log}}_{2}4$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}2{\mathrm{log}}_{2}{\mathrm{log}}_{2}x+{\mathrm{log}}_{\frac{1}{2}}{\mathrm{log}}_{2}(2\sqrt{2}x)={\mathrm{log}}_{2}{\mathrm{log}}_{2}4$**$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{2}{\mathrm{log}}_{2}{x}^{2}-{\mathrm{log}}_{2}{\mathrm{log}}_{2}(2\sqrt{2}x)={\mathrm{log}}_{2}{\mathrm{log}}_{2}4$**

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{2}\frac{{\mathrm{log}}_{2}{x}^{2}}{{\mathrm{log}}_{2}(2\sqrt{2}x)}={\mathrm{log}}_{2}{\mathrm{log}}_{2}4$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{{\mathrm{log}}_{2}{x}^{2}}{{\mathrm{log}}_{2}(2\sqrt{2}x)}={\mathrm{log}}_{2}4$

Please suggest whether is it the right approach... thanks...

$2{\mathrm{log}}_{2}{\mathrm{log}}_{2}x+{\mathrm{log}}_{\frac{1}{2}}{\mathrm{log}}_{2}(2\sqrt{2}x)=1$

My approach :

Solution : Here right hand side is constant term so convert it into log of same base as L.H.S. therefore, 1 can be written as ${\mathrm{log}}_{2}{\mathrm{log}}_{2}4$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}2{\mathrm{log}}_{2}{\mathrm{log}}_{2}x+{\mathrm{log}}_{\frac{1}{2}}{\mathrm{log}}_{2}(2\sqrt{2}x)={\mathrm{log}}_{2}{\mathrm{log}}_{2}4$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{2}\frac{{\mathrm{log}}_{2}{x}^{2}}{{\mathrm{log}}_{2}(2\sqrt{2}x)}={\mathrm{log}}_{2}{\mathrm{log}}_{2}4$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{{\mathrm{log}}_{2}{x}^{2}}{{\mathrm{log}}_{2}(2\sqrt{2}x)}={\mathrm{log}}_{2}4$

Please suggest whether is it the right approach... thanks...

asked 2022-05-28

Lower bound for logarithm?

Given a real $c$ such that $1<c$, is there any known and direct lower bound, other than $0$, for $(\mathrm{ln}c)$, i.e., $A<\mathrm{ln}c$

Thanks

Given a real $c$ such that $1<c$, is there any known and direct lower bound, other than $0$, for $(\mathrm{ln}c)$, i.e., $A<\mathrm{ln}c$

Thanks