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I have the following exponential equation:


I use the product rule to separate the exponent from the base and that resulted in:

$$2x \cdot \ln(3) + x \cdot \ln(3) = 2$$

Then I factored out the $x$ from the two terms and divided the non-x term onto the other side to get:


Although this doesn't seem to work when I check my solution by plugging the value back into the original equation. Can anyone point out where I made a mistake?

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You are calculating (sort of) as if $\ln(a+b)=\ln a+\ln b$. This is not true. – André Nicolas Feb 25 '14 at 0:51
Use $y = 3^{x}$. – Felix Marin Feb 25 '14 at 1:37
up vote 3 down vote accepted

Whenever you see something like $a^{2x}+b^x+c=0$, it can usually be solved by factoring by substitution. In your case, you can use factoring. Let $3^x=y$. So, $3^{2x}=(3^x)^2=y^2$. The equation now becomes $y^2+y-2=0$. This factors to $(y-1)(y+2)=0$. So, the solutions are $y = -2, \ 1$. That means that $3^x=-2, \ 1$, because of substituting $y$ as $3^x$ again. Right?


How can any exponential in the form $a^x$ ($a \ge 0$) have a negative answer? The least that it can get to is $0$. How, then, can $3^x=-2$? It is impossible! So that is why we have to set a restriction that $3^x \ge 0$. So, $-2$ is not a solution. That leaves $3^x=1$, which is easy to solve. $$3^x=1$$ $$\ln(3^x)=\ln(1)$$ $$x\ln(3)=\ln(1)$$ $$\displaystyle x=\dfrac{\ln(1)}{\ln(3)}$$ Remember that $\ln(1)=0$. So the solution is $x=0$.

Therefore there is one solution, which is: $$\displaystyle \boxed{x=0}$$ I hope I helped you out!

P.S. I do not want to sound like a guy that goes crazy about Yin and Yang, but I want you to know that what you see above is the beauty of mathematics; why so many people love math. For almost every question, there is more than one way to get to the answer. I could have just immediately said that $x=0$, because $a^0 = 1, \ a\in \mathbb R$. But another way to do it is using the logarithms. There just might be another way other than factoring the equation in the first step as well! Such is the beauty of mathematics....

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The solution aren't $-1$ and $2$. They are $-2$ and $1$. And the final solution is $x=0$ – Stefan4024 Feb 25 '14 at 1:13
@Stefan4024 How is the final solution $x=0$? Simply plugging in that value gives you $-2=0$, which is definitely false! – JChau Feb 25 '14 at 1:17
$3^0 =$ what? Look it other way $\ln(1) = ...$ – Stefan4024 Feb 25 '14 at 1:18
Wouldn't it be $3^{2 \cdot 0} + 3^{0} -2 = 0$. Which would be $1 + 1 -2=0$, which indeed is $0=0$? I believe he is correct. – ProfessorStealth Feb 25 '14 at 1:20
I must ask though, is there another way to solve this? My professor will probably want me to do it a longer route without the clever substitution. :( – ProfessorStealth Feb 25 '14 at 1:22

Substitute $3^x = y$ and $3^{2x} = y^2$ and see what happens...

Be careful! $3^x = y > 0$

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Letting $y=3^x$ and reaching $(y+2)(y-1)=0$ is precisely how your professor will expect you to do it.

There is another way. It is clear that $x=0$ is a solution. To see there are no others, note that if $x\lt 0$ then $3^{2x}$ and $3^x$ are each less than $1$, while if $x\gt 0$ then each of $3^{2x}$ and $3^x$ is bigger than $1$. So there are no solutions other than $0$.

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What if $0 < x < 1$? – JChau Feb 25 '14 at 1:39
Then $3^{2x}\gt 1$ and $3^x\gt 1$. The function $3^t$ is an increasing function. – André Nicolas Feb 25 '14 at 1:51

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