I have trouble solving this:
$$3^x+3^{2-x}=8$$
I have tried substituting $3^x=z$ but that doesn't seem to help much.
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4Substituting $3^x=z$ ought to give you $z+9/z=8$. That looks quite helpful. Multiply that by $z$ and you have a quadratic equation. – hmakholm left over Monica Jul 22 '16 at 09:24
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I did get to the quadratic but I cant find $x$ after because I can't take the logarithm. I get a square root in the solution – DoubleOseven Jul 22 '16 at 09:25
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It has 2 real solutions. This equation can be written as $9^x - 8\times 3^x + 9 = 0$ which is now a quadratic in $3^x$ – Shailesh Jul 22 '16 at 09:26
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@DenisDüsseldorf Assuming $\log (a+b)=\log(a)+\log(b)$, you are correct. Sadly, that's not the case. – 5xum Jul 22 '16 at 09:26
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2You know, Double, you'd get much more helpful answers if you'd tell people in the first place how far you've gotten and where you get stuck. – Gerry Myerson Jul 22 '16 at 09:26
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1@DoubleOseven: By "I can't take the logarithm" do you mean that the quadratic has no real roots, or only negative roots? If that is the case, the original equation has no (real) solutions. But in fact is has (unless I'm doing something wrong) two real roots. – hmakholm left over Monica Jul 22 '16 at 09:27
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@HenningMakholm I mean I can't simplify the equation to get to the solution – DoubleOseven Jul 22 '16 at 09:28
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1@DoubleOseven: What does that mean? Which roots do you get for the quadratic? – hmakholm left over Monica Jul 22 '16 at 09:29
3 Answers
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Hint:
Let me just rewrite the equation for you
$$3^x + 3^{2-x} = 8\\ 3^x + 3^2\cdot 3^{-x} = 8\\ 3^x + 9\cdot (3^x)^{-1} = 8.$$
Now, try the substitution again. What do you get?
5xum
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That is what I did but if you continue you will see that the solution can't be simplified. Am I wrong? – DoubleOseven Jul 22 '16 at 09:29
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2@DoubleOseven: When I continue there seems to be absolutely no problems with getting the two solutions for $x$. Instead of just continuing to assert that there is some problem that you're stuck at, you need to show exactly what point you're stuck at instead of leaving it to the reader to guess what you're doing. – hmakholm left over Monica Jul 22 '16 at 09:31
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@DoubleOseven Yes. Please write down the equation you get, and the solution. – 5xum Jul 22 '16 at 09:31
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5@DoubleOseven Also, in future, please don't just say "that's not possible". Instead say "I tried this, got this far, now I don't know how to proceed". If you want good answers, show your work. That way, we can address the part that is troubling you directly. – 5xum Jul 22 '16 at 09:32
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Multiply $3^x$ to both sides: $3^{2x} + 9 = 8\cdot 3^x \implies 3^{2x} - 8\cdot 3^x + 9 = 0\implies (3^x - 4)^2 = 7 \implies 3^x - 4 = \pm \sqrt{7}\implies 3^x = 4 \pm \sqrt{7} \implies x = \log_{3}(4 \pm \sqrt{7})$.
DeepSea
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You can also rewrite it as $$ 3^{2x} - 8\cdot3^x + 3^2 = 0 $$
given that $3^{-x} \ne 0$. Then, solving for $3^x$ gives you
$$ 3^x = \frac{4 \pm \sqrt{16 - 9}}{1}. $$
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3@DoubleOseven: Haven't you learned how to solve an equation of the form $a^x=b$? You solve it by taking the base-$a$ logarithm on both sides! – hmakholm left over Monica Jul 22 '16 at 09:32