 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
${10}^{\mathrm{log}\left(x\right)}=x$
Thanks Genesis Rosario

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
${b}^{{\mathrm{log}}_{b}\left(x\right)}=x$
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 $z=x{e}^{x}$
$z=x{e}^{x}⟺W\left(z\right)=x$
Since you're curious, here are the solutions
$2={3}^{x}x=x{3}^{x}=x{e}^{\mathrm{ln}{3}^{x}}=x{e}^{x\mathrm{ln}3}$
$2\mathrm{ln}3=\left(x\mathrm{ln}3\right){e}^{x\mathrm{ln}3}$
$W\left(2\mathrm{ln}3\right)=x\mathrm{ln}3$
$x=\frac{W\left(2\mathrm{ln}3\right)}{\mathrm{ln}3}$
And after some trial and error,
$2={3}^{x}+x$
$2-x={3}^{x}$
$\left(2-x\right){3}^{2}={3}^{x}{3}^{2}$
$\left(2-x\right)\frac{{3}^{2}}{{3}^{x}}=9$
$9=\left(2-x\right)\frac{{3}^{2}}{{3}^{x}}=\left(2-x\right){3}^{2-x}=\left(2-x\right){e}^{\mathrm{ln}{3}^{2-x}}=\left(2-x\right){e}^{\left(2-x\right)\mathrm{ln}3}$
$9\mathrm{ln}3=\left(\left(2-x\right)\mathrm{ln}3\right){e}^{\left(2-x\right)\mathrm{ln}3}$
$W\left(9\mathrm{ln}3\right)=\left(2-x\right)\mathrm{ln}3=2\mathrm{ln}3-x\mathrm{ln}3$
$W\left(9\mathrm{ln}3\right)-2\mathrm{ln}3=-x\mathrm{ln}3$
$x=\frac{2\mathrm{ln}3-W\left(9\mathrm{ln}3\right)}{\mathrm{ln}3}$

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