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How to solve this logarithm equation? $$\frac12\cdot[\log(x) + \log(2)) + \log[\sqrt{2x} + 1] = \log(6).$$

The answer is $2$.

I've tried to solve it, but I don't know how to proceed:

$\frac12\log(2x) + \log[\sqrt{2x} + 1] = \log(6)$

$\log([(2x)^\frac12] + \log[\sqrt{2x} + 1] = \log(6)$

$\log[\sqrt{2x}] + \log[\sqrt{2x} + 1] = \log(6)$

$\log(\sqrt{2x}\cdot [\sqrt{2x} + 1]) = \log(6)$

$\sqrt{2x}\cdot[\sqrt{2x} + 1] = 6$

$2x + \sqrt{2x} = 6$

$2x + \sqrt{2}\sqrt{x} - 6 = 0$

Math Lover
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user580053
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  • Hint: Let $t=\sqrt{2x}$ – Pizzaroot Jul 28 '18 at 01:12
  • From your final step, make a change of variable. Let's use instead the variable $u$ where $u=\sqrt{x}$. Your last line rewritten then with $u$ is $2u^2+\sqrt{2}\cdot u - 6=0$. This is a quadratic equation which you should know how to handle. Make sure that at the end, you check to make sure that any solutions you might have gotten make sense throughout the earlier steps and in the change-of-variable (i.e. discard any negative values of $u$ and/or negative values of $x$). – JMoravitz Jul 28 '18 at 01:12

2 Answers2

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$$\frac 12 [\log(x)+\log(2)]+\log(\sqrt{2x}+1)=\log(6)$$ By $\log(a^n)=n\log(a)$, we get: $$\log(\sqrt x)+\log(\sqrt 2)+\log(\sqrt{2x}+1)=\log(6)$$ By $\log(p)+\log(q)=\log(pq)$, we get: $$\log(\sqrt x\cdot\sqrt 2\cdot(\sqrt{2x}+1))=\log(6)$$ $$\log(2x+\sqrt{2x})=\log(6)$$

$$\to 2x+\sqrt{2x}=6$$

One can then iterate: $$x_{n+1}=\frac 12(6-\sqrt{2x_n}); x_0=1$$ and show this converges to $x=2$, testing this value shows it as correct.

Rhys Hughes
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  • Why would you define this sequence and show it converges, when you can elementary solve the equations easily? I like the idea tho. – Cornman Jul 28 '18 at 01:21
  • Convergence need only be demonstrated by calculator (as I would do it), then proven with an insertion of $x=2$ into the original question. – Rhys Hughes Jul 28 '18 at 01:22
  • It is just my personal preference to iterate rather than embark on a sometimes convoluted journey through conventional methods. Also it eliminates false solutions which may occur – Rhys Hughes Jul 28 '18 at 01:24
  • And what do you do, if there are more solutions than one? – Cornman Jul 28 '18 at 01:27
  • There are always two ways to iterate an equation. One will lead to solution, another a different solution. Here the other iterate would be: $$x_{n+1}=\frac12(6-2x_n)^2,x_0=1$$. It's rather simple here to see that this iterate diverges at a very rapid pace by calculation, and so there is no second solution. – Rhys Hughes Jul 28 '18 at 01:31
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Square both sides of the equation $\sqrt{2x}=6-2x$ to obtain $2x=36-24x+4x^2$ which will give you this quadratic equation $4x^2-26x+36=0$. If we simplify the equation further by dividing by the common factor $2$ we obtain $2x^2-13x+18=0$. This should be easy to solve if you use the quadratic formula or by factorizing the left side and finding its roots.

user573497
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  • Squaring both sides is tricky, since taking the square is not an equivalent transformation. Note, that the RHS can be negativ. This gives "fake solutions". The better way is to subsitute. Or using the quadratic formula. – Cornman Jul 28 '18 at 01:17
  • One last comment: you must test the $x$ values into the original equation to see if they satisfies the original equation. – user573497 Jul 28 '18 at 01:17
  • @Cornman you are right, but he can still use the quadratic formula and then dicard any fake solution of the original equation. – user573497 Jul 28 '18 at 01:19
  • Just for info: $4x^2-26x+36 = (x-2)(4x-18)$. – Math Lover Jul 28 '18 at 01:59
  • Did you seriously not notice that $4, 26$ and $36$ are all even so you actually have $$2x^2-13x+18=0$$. – Rhys Hughes Jul 28 '18 at 02:12
  • @RhysHughes I editted the answer. Thank you. – user573497 Jul 28 '18 at 02:21