Can one use logarithms to solve the equations $2={3}^{x}+x$ and $2={3}^{x}x$?

I can only solve halfway through.

And why is

${10}^{\mathrm{log}(x)}=x$

Thanks

I can only solve halfway through.

And why is

${10}^{\mathrm{log}(x)}=x$

Thanks

HypeMyday3m
2022-09-25
Answered

Can one use logarithms to solve the equations $2={3}^{x}+x$ and $2={3}^{x}x$?

I can only solve halfway through.

And why is

${10}^{\mathrm{log}(x)}=x$

Thanks

I can only solve halfway through.

And why is

${10}^{\mathrm{log}(x)}=x$

Thanks

You can still ask an expert for help

Genesis Rosario

Answered 2022-09-26
Author has **11** answers

By definition, a log is a quantity representing the power to which a fixed number (the base) must be raised to produce a given number.

Below is a simple example,

${10}^{2}=100$

So

${\mathrm{log}}_{10}100={\mathrm{log}}_{10}{10}^{2}=2{\mathrm{log}}_{10}10=2$

And

${10}^{{\mathrm{log}}_{10}100}={10}^{2}=100$

Generally

${b}^{{\mathrm{log}}_{b}(x)}=x$

Also, the solutions to both of those problems cannot be found using elementary functions.

To solve these equations, we must use the Lambert W function. This function will provide the value of $x$ in equations that take the form $z=x{e}^{x}$

$z=x{e}^{x}\u27faW(z)=x$

Since you're curious, here are the solutions

$2={3}^{x}x=x{3}^{x}=x{e}^{\mathrm{ln}{3}^{x}}=x{e}^{x\mathrm{ln}3}$

$2\mathrm{ln}3=(x\mathrm{ln}3){e}^{x\mathrm{ln}3}$

$W(2\mathrm{ln}3)=x\mathrm{ln}3$

$x=\frac{W(2\mathrm{ln}3)}{\mathrm{ln}3}$

And after some trial and error,

$2={3}^{x}+x$

$2-x={3}^{x}$

$(2-x){3}^{2}={3}^{x}{3}^{2}$

$(2-x)\frac{{3}^{2}}{{3}^{x}}=9$

$9=(2-x)\frac{{3}^{2}}{{3}^{x}}=(2-x){3}^{2-x}=(2-x){e}^{\mathrm{ln}{3}^{2-x}}=(2-x){e}^{(2-x)\mathrm{ln}3}$

$9\mathrm{ln}3=((2-x)\mathrm{ln}3){e}^{(2-x)\mathrm{ln}3}$

$W(9\mathrm{ln}3)=(2-x)\mathrm{ln}3=2\mathrm{ln}3-x\mathrm{ln}3$

$W(9\mathrm{ln}3)-2\mathrm{ln}3=-x\mathrm{ln}3$

$x=\frac{2\mathrm{ln}3-W(9\mathrm{ln}3)}{\mathrm{ln}3}$

Below is a simple example,

${10}^{2}=100$

So

${\mathrm{log}}_{10}100={\mathrm{log}}_{10}{10}^{2}=2{\mathrm{log}}_{10}10=2$

And

${10}^{{\mathrm{log}}_{10}100}={10}^{2}=100$

Generally

${b}^{{\mathrm{log}}_{b}(x)}=x$

Also, the solutions to both of those problems cannot be found using elementary functions.

To solve these equations, we must use the Lambert W function. This function will provide the value of $x$ in equations that take the form $z=x{e}^{x}$

$z=x{e}^{x}\u27faW(z)=x$

Since you're curious, here are the solutions

$2={3}^{x}x=x{3}^{x}=x{e}^{\mathrm{ln}{3}^{x}}=x{e}^{x\mathrm{ln}3}$

$2\mathrm{ln}3=(x\mathrm{ln}3){e}^{x\mathrm{ln}3}$

$W(2\mathrm{ln}3)=x\mathrm{ln}3$

$x=\frac{W(2\mathrm{ln}3)}{\mathrm{ln}3}$

And after some trial and error,

$2={3}^{x}+x$

$2-x={3}^{x}$

$(2-x){3}^{2}={3}^{x}{3}^{2}$

$(2-x)\frac{{3}^{2}}{{3}^{x}}=9$

$9=(2-x)\frac{{3}^{2}}{{3}^{x}}=(2-x){3}^{2-x}=(2-x){e}^{\mathrm{ln}{3}^{2-x}}=(2-x){e}^{(2-x)\mathrm{ln}3}$

$9\mathrm{ln}3=((2-x)\mathrm{ln}3){e}^{(2-x)\mathrm{ln}3}$

$W(9\mathrm{ln}3)=(2-x)\mathrm{ln}3=2\mathrm{ln}3-x\mathrm{ln}3$

$W(9\mathrm{ln}3)-2\mathrm{ln}3=-x\mathrm{ln}3$

$x=\frac{2\mathrm{ln}3-W(9\mathrm{ln}3)}{\mathrm{ln}3}$

asked 2022-04-08

Use properties of logarithms to condense the logarithmic expression . Write the expression as a single logarithm whose coefficient is 1. Where possible, evaluate logarithmic expressions without using a calculator.

$3\mathrm{ln}x-\frac{1}{3}\mathrm{ln}y$

asked 2021-02-18

Solve the equations and inequalities. Write the solution sets to the inequalities in interval notation.

asked 2021-12-13

How do you solve $\mathrm{ln}x=2?$

asked 2022-08-13

How do I find the inverse of this exponential function?

$x=-3({3}^{-x})+9$

I know the steps up until a certain point.

$x=-3({3}^{-y})+9$

$x-9=-3({3}^{-y})$

$\frac{(x-9)}{-3}={3}^{y}$

$ln(\frac{x-9}{-3})=-y\ast ln3$

Not sure what to do from here. I know I have to get y by itself but thats it. Can anyone help please?

$x=-3({3}^{-x})+9$

I know the steps up until a certain point.

$x=-3({3}^{-y})+9$

$x-9=-3({3}^{-y})$

$\frac{(x-9)}{-3}={3}^{y}$

$ln(\frac{x-9}{-3})=-y\ast ln3$

Not sure what to do from here. I know I have to get y by itself but thats it. Can anyone help please?

asked 2022-03-30

I am new to logarithms and I am having trouble with this logarithm system.

${\mathrm{log}}_{9}\left(x\right)+{\mathrm{log}}_{y}\left(8\right)=2$

$\mathrm{log}}_{x}\left(9\right)+{\mathrm{log}}_{8}\left(y\right)=\frac{8}{3$

A step-by-step procedure would be highly appreciated.

Thanks in advance.

A step-by-step procedure would be highly appreciated.

Thanks in advance.

asked 2022-10-03

lHopitals $\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule{thickmathspace}{0ex}}(\mathrm{ln}x{)}^{3x}$?

Okay, so what do I do with that power? I need to rewrite the term as fractions. How?

If it was the inner function that's in the power of something: $\mathrm{ln}{x}^{\frac{1}{3x}}$ then I'd just simply rewritten it as $\frac{1}{3x}\cdot \mathrm{ln}x=\frac{\mathrm{ln}x}{3x}$

Okay, so what do I do with that power? I need to rewrite the term as fractions. How?

If it was the inner function that's in the power of something: $\mathrm{ln}{x}^{\frac{1}{3x}}$ then I'd just simply rewritten it as $\frac{1}{3x}\cdot \mathrm{ln}x=\frac{\mathrm{ln}x}{3x}$

asked 2022-10-14

Help with Evaluating a Logarithm

A precalculus text asks us to evaluate ${\mathrm{log}}_{8}{\displaystyle \frac{\sqrt{2}\cdot \sqrt[3]{256}}{\sqrt[6]{32}}}$

I do the following: ${\mathrm{log}}_{8}{\displaystyle \frac{\sqrt{2}\cdot \sqrt[3]{({2}^{2}{)}^{3}\cdot {2}^{2}}}{\sqrt[6]{{2}^{3}\cdot {2}^{2}}}}$

$\equiv {\mathrm{log}}_{8}{\displaystyle \frac{\sqrt{2}\cdot {2}^{2}\cdot \sqrt[3]{{2}^{2}}}{\sqrt{2}\cdot \sqrt[6]{{2}^{2}}}}$

$\equiv {\mathrm{log}}_{8}{\displaystyle \frac{{2}^{2}\cdot \sqrt[3]{{2}^{2}}}{\sqrt[3]{2}}}$

and then I'm stumped.

Hints?

A precalculus text asks us to evaluate ${\mathrm{log}}_{8}{\displaystyle \frac{\sqrt{2}\cdot \sqrt[3]{256}}{\sqrt[6]{32}}}$

I do the following: ${\mathrm{log}}_{8}{\displaystyle \frac{\sqrt{2}\cdot \sqrt[3]{({2}^{2}{)}^{3}\cdot {2}^{2}}}{\sqrt[6]{{2}^{3}\cdot {2}^{2}}}}$

$\equiv {\mathrm{log}}_{8}{\displaystyle \frac{\sqrt{2}\cdot {2}^{2}\cdot \sqrt[3]{{2}^{2}}}{\sqrt{2}\cdot \sqrt[6]{{2}^{2}}}}$

$\equiv {\mathrm{log}}_{8}{\displaystyle \frac{{2}^{2}\cdot \sqrt[3]{{2}^{2}}}{\sqrt[3]{2}}}$

and then I'm stumped.

Hints?