HypeMyday3m

2022-09-25

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

Genesis Rosario

Beginner2022-09-26Added 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,

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

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

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

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

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

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

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

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

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

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