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

Question

Could someone explain how would you solve:

$$2=3^x + x$$

and

$$2=3^x \cdot x$$

I can only solve halfway through.

And why is

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

Thanks

Answer 1

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 \[ \log_{10} 100=\log_{10} 10^{2}=2\log_{10} 10= 2 \] And \[ 10^{\log_{10} 100}=10^{2} = 100 \] Generally \[ b^{\log_{b}(x)}=x \] 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 $z=xe^{x}$. \[ z=xe^{x}\Longleftrightarrow W(z)=x \] Since you're curious, here are the solutions \[ 2=3^{x}x=x3^{x}=xe^{\ln 3^{x}}=xe^{x\ln 3} \] \[ 2\ln 3=(x\ln 3)e^{x\ln 3} \] \[ W(2\ln 3) = x\ln 3 \] \[ x = \frac{W(2\ln 3)}{\ln 3} \] And after some trial and error, \[ 2=3^{x}+x \] \[ 2-x=3^{x} \] \[ (2-x)3^{2}=3^{x}3^{2} \] \[ (2-x)\frac{3^{2}}{3^{x}}=9 \] \[ 9=(2-x)\frac{3^{2}}{3^{x}}=(2-x)3^{2-x}=(2-x)e^{\ln 3^{2-x}}=(2-x)e^{(2-x)\ln 3} \] \[ 9\ln 3=((2-x)\ln 3)e^{(2-x)\ln 3} \] \[ W(9\ln 3)=(2-x)\ln 3 = 2\ln 3-x\ln 3 \] \[ W(9\ln 3) - 2\ln 3= -x\ln 3 \] \[ x=\frac{2\ln 3 - W(9\ln 3)}{\ln 3} \]

Answer 2

The two first equations cannot be solved using elementary functions.

There are algebraic ways to explain the third equation, but here is an intuitive way:

What is the definition of $\log_{10} x$? It means: to what power do I have to raise $10$ to get $x$? Imagine you raise $10$ to that particular power. What do you get? $x$!