Solve for x:

${5}^{3x+24}=1$

${5}^{3x+24}=1$

Luz Stokes
2022-07-27
Answered

Solve for x:

${5}^{3x+24}=1$

${5}^{3x+24}=1$

You can still ask an expert for help

Bradley Sherman

Answered 2022-07-28
Author has **17** answers

Consider the equation ${5}^{3x+24}=1$

Taking natural logarithms bot sides:

$\mathrm{ln}{5}^{3x+24}=\mathrm{ln}1\phantom{\rule{0ex}{0ex}}\mathrm{ln}{5}^{3x+24}=0\phantom{\rule{0ex}{0ex}}(3x+24)\mathrm{ln}5=0\phantom{\rule{0ex}{0ex}}3x+24=0\phantom{\rule{0ex}{0ex}}3x=-24\phantom{\rule{0ex}{0ex}}x=\frac{-24}{3}\phantom{\rule{0ex}{0ex}}x=-8$

Therefore, the solution is x=-8

Taking natural logarithms bot sides:

$\mathrm{ln}{5}^{3x+24}=\mathrm{ln}1\phantom{\rule{0ex}{0ex}}\mathrm{ln}{5}^{3x+24}=0\phantom{\rule{0ex}{0ex}}(3x+24)\mathrm{ln}5=0\phantom{\rule{0ex}{0ex}}3x+24=0\phantom{\rule{0ex}{0ex}}3x=-24\phantom{\rule{0ex}{0ex}}x=\frac{-24}{3}\phantom{\rule{0ex}{0ex}}x=-8$

Therefore, the solution is x=-8

asked 2022-08-02

If $\mathrm{log}$ sub(a)10 = 0.250, then $\mathrm{log}$ sub(10)a equals ?

Breakdown:

$\mathrm{log}$ sub(a)10 = 0.250

The solution requires me to the take the base(a) antilogarithmof both sides. That would be

$10={a}^{0.250}$.

Breakdown:

$\mathrm{log}$ sub(a)10 = 0.250

The solution requires me to the take the base(a) antilogarithmof both sides. That would be

$10={a}^{0.250}$.

asked 2022-09-04

Rewrite as a single log with a coefficient of one.

$(4\mathrm{ln}(x))-(3\mathrm{ln}(x))+(1/4\mathrm{ln}(x))$

$(4\mathrm{ln}(x))-(3\mathrm{ln}(x))+(1/4\mathrm{ln}(x))$

asked 2021-10-18

Use The Properties Of Logarithms To Simplify

$\mathrm{log}}_{23.2}-{\mathrm{log}}_{20.025$

asked 2022-05-21

Simplifying an expression using a logarithm

I have the following expression

$\frac{1}{1+\rho}(1+n{)}^{(1-\sigma )}\ast (1+{\gamma}_{A}{)}^{1-\sigma}<1$

and have to use logarithms to get the following

$(1-\sigma )(n+{\gamma}_{A})<\rho $

Could someone please offer me some help on how to do it? That's a homework question in Macroeconomics - we got the solutions and were told to use the logarithms to get them. However, I'm struggling with the complete procedure, so any help is more than welcome.

Thanks!

I have the following expression

$\frac{1}{1+\rho}(1+n{)}^{(1-\sigma )}\ast (1+{\gamma}_{A}{)}^{1-\sigma}<1$

and have to use logarithms to get the following

$(1-\sigma )(n+{\gamma}_{A})<\rho $

Could someone please offer me some help on how to do it? That's a homework question in Macroeconomics - we got the solutions and were told to use the logarithms to get them. However, I'm struggling with the complete procedure, so any help is more than welcome.

Thanks!

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 2022-05-14

Asymptotics of logarithms of functions

If I know that $\underset{x\to \mathrm{\infty}}{lim}{\displaystyle \frac{f(x)}{g(x)}}=1$, does it follow that $\underset{x\to \mathrm{\infty}}{lim}{\displaystyle \frac{\mathrm{log}f(x)}{\mathrm{log}g(x)}}=1$ as well? I see that this definitely doesn't hold for $\frac{{e}^{f(x)}}{{e}^{g(x)}}$ (take $f(x)=x+1$ and $g(x)=x$), but I'm not sure how to handle the other direction.

If I know that $\underset{x\to \mathrm{\infty}}{lim}{\displaystyle \frac{f(x)}{g(x)}}=1$, does it follow that $\underset{x\to \mathrm{\infty}}{lim}{\displaystyle \frac{\mathrm{log}f(x)}{\mathrm{log}g(x)}}=1$ as well? I see that this definitely doesn't hold for $\frac{{e}^{f(x)}}{{e}^{g(x)}}$ (take $f(x)=x+1$ and $g(x)=x$), but I'm not sure how to handle the other direction.

asked 2022-05-19

simplify the following logarithm and roots expression

please , I need help to solve this problem Can anyone solve it simplify the following

$\mathrm{log}\sqrt[4]{729\sqrt[3]{{\displaystyle \frac{1}{39}}\sqrt{{\left({\displaystyle \frac{1}{27}}\right)}^{4}}}}$

thank in advance to all

please , I need help to solve this problem Can anyone solve it simplify the following

$\mathrm{log}\sqrt[4]{729\sqrt[3]{{\displaystyle \frac{1}{39}}\sqrt{{\left({\displaystyle \frac{1}{27}}\right)}^{4}}}}$

thank in advance to all