I need to solve the exponential equation $((x + 4)10^x)/(x - 3) = 2x(10^x)$ assuming the fact that $2^x$ is always positive.

The example uses the case $x^3-2^x - 3(2^x) = 0$ factors out $2^x$ leaving $2^x (x^3-3) = 0$. They then apply the zero-product property and solve $2^x$ and $x^3 - 3 = 0$. Does this still apply for my more complex question? In that case, would I get one side of the equation to $0$, then apply this same property? Thanks in advance for any assistance.

  • $\begingroup$ Multiply both sides of the equation by $x-3$, and transpose everything to the left. Factor and note what you have... $\endgroup$ – J. M. is a poor mathematician Nov 22 '11 at 17:18
  • $\begingroup$ Since $10^x$ is not equal to $0$, you can divide both sides by $10^x$ to get an equivalent equation. By the way, there is (currently) a typo in the explanatory material that involves $2^x$. $\endgroup$ – André Nicolas Nov 22 '11 at 17:24
  • $\begingroup$ That was probably my typo, but was just an example of the procedure. $\endgroup$ – erimar77 Nov 22 '11 at 17:26

$$\frac{(x+4)10^x-2x(x-3)10^x}{x-3}=0 \Rightarrow \frac{10^x(-2x^2+7x+4)}{x-3}=0 \Rightarrow$$

$ \Rightarrow-2x^2+7x+4=0$ and $x\neq 3$


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