# Rules for natural logging both sides of an equation?

I just had a question about the rules for natural logs and I'm not too sure how to word my question into google to get the answers I'm seeking so here it goes I guess. Specifically problems like this.

$$1.006^{\left(60-x\right)}+\left(2\cdot 1.006^{\left(60-2x\right)}\right)=3.823$$

For the equation mentioned above, normally I would natural log both sides to move the exponents down so to speak. But what's tripping me up is the "2". I'm not sure how natural logging both sides would work with the 2 being there.

$$(60-X)ln(1.006)+(60-2X)(ln(2))(ln(1.006))=ln(3.823)$$

This is what I think of naturally probably because I don't know all of the natural log rules. And I was wondering what the actual rule is and how the "2" works in this mess. I'm probably just too tired to think this through but I can't sleep right tonight if I don't understand it. I guess that's about it. Thanks in advance.

• You might want to bear in mind that you're assuming $\ln(a+b) = \ln(a) + \ln(b)$ in your work in the equation. That's not actually true in general; you're mixing it up with $\ln(ab) = \ln(a) + \ln(b)$. May 30 '20 at 5:31
• What you want to do instead is to let $y=1.006^{-x}$ and then you get a quadratic equation for $y$. May 30 '20 at 5:33
• Have you tried out my suggestion, Ross? May 31 '20 at 12:09

Your second line is incorrect because $$ln(a+b)\neq{ln(a)+ln(b)}$$
But you can use the law of exponents to rewrite the equation this way. $$1.006^{60}1.006^{-x}+ 2\left(1.006^{60}1.006^{-2x}\right)=3.823$$
And you can define another variable as a comment suggests. $$y=1.006^{-x}$$
After substitution, we have $$1.006^{60}y+ 2\left(1.006^{60}y^2\right)=3.823$$ Use the quadratic formula to solve for y and then use those solutions to solve for x.