Need suggestion to solving logarithm equation of $ \log(\log(x+3))+\log(2) =\log(\log(16x)) $

Question

$$ \log(\log(x+3))+\log(2) =\log(\log(16x)) $$ My work so far is :

Step 1: I using the property of $\log(a)+\log(b)=\log(a\times b)$. Where $a=\log(x+3)$ and $ b = 2$ $$ \log(2\times \log(x+3))=\log(\log(16x)) $$ Step 2 : I applying anti logarithm. $$2 \times \log(x+3)=\log(16x)$$ Step 3 : I using the property of $b \times \log(c)=\log(c^b)$. Where $b=2$ and $c=(x+3)$ $$ \log((x+3)^2)=\log(16x) $$ Step 4 : I applying anti logarithm. $$(x+3)^2=16x $$ Step 5 : I expanding the exponent and construct the quadratic form. $$ x^2+6x+9-16x=0$$ $$ x^2-10x+9=0 $$ Step 6 : I factoring the quadratic form to get two solutions. $$ (x-1)(x-9)=0 $$ $$ x_1 = 1 \lor x_2= 9$$ Step 7 : I finding the definition term for $\log(\log(x+3))$. $$ \log(x+3)>0 $$ $$ \log(x+3)>\log(1) $$ $$x+3>1$$ $$x>-2$$ Step 8 : I finding the definition term for $\log(x+3)$. $$x+3>0$$ $$x>-3$$ Step 9 : I finding the definition term for $\log(\log(16x))$. $$ \log(16x)>0$$ $$ \log(16x)>\log(1)$$ $$16x>1$$ $$x>\frac{1}{16}$$ Step 10 : I finding the definition term for $\log(16x)$. $$16x>0$$ $$x>0$$ Step 11 : I making a summary for all definition term. $$x>\frac{1}{16}$$ Step 12 : I filtering the solutions of equations using the definition term to found the set of solutions $$x= \{1,9\}$$

Am i do the correct solution for equation above ? I need suggestion and correction if i make a mistake and forget about logarithm property. Thanks for your help,sir.

Answer 1

Yes, your derivation of $x=1$ and $x=9$ is excellent.

Answer 2

Yes we have

$$\log(\log(x+3))+\log(2) =\log(\log(16x))\iff x^2-10x+9=0$$

with the conditions

• $x+3>1$
• $16x>1$

that is $x>\frac1{16}$ and thus $x= \{1,9\}$.

Answer 3

You are correct. Personally, I would do the following:

Using logarithmic properties, we have

$\log(\log(x+3)) + \log2 = \log(\log(x+3)\cdot 2).$ We also have by raising $10$ to the power of both sides that $\log(x+3)\cdot 2=\log(16x).$ Further simplification yields $(x+3)^2=16x.$ And solving gives $x=1$ or $x=9.$ We can plug these values back into the original equation to check if they work, and they do because they're positive.