# Can the following equations be solved without the need of numerical methods?

I'm taking advanced algebra in school. I have been asked to solve two equations:

$\log_{6}(1-x) + \log(x^{2}-9) = 2 \\$

$3^{x+2} + 2^x = 5$

The teacher said this equations can be solved analytically. However, I've been trying vary methods (change of base in the first one and then trying to put all the expressions with X in one side but I get a term powered to a logarithm constant (it's pretty straightforward to get this). To be a bit more precise, I do the change of basis of the first term and after a few algebra I arrive to the expression:

$(1-x)(x^2-9)^{log(6)} = 10^{2log(6)} \\$

Which I can't manage to solve, and to be quite honest I'm not completely sure if that is the right path to solve the question.

In the other equation I really don't know even where to start. I tried to separate 5 as 2 + 3, then pass each term to one side, so that I get sums of powers of 2 on one side and powers of 3 on the other side. However, the remaining expression is even worse than the one I got from the previous equation.

I wanted to know if you can manage to solve this and if you can, could you please give me a hint since I think I'm about to give up, I've tried everything I think but nothing seems to solve this equations.

• I'd try adding another variable and getting rid of the logs: 6^y=1-x and 2-10^y=x^2-9 Now you have two equations with two variables. – theDoctor Oct 27 '17 at 6:40

For the first, Alpha again doesn't. I didn't know if the second log is base $10$ or base $e$, but it doesn't find one either way. For equations of this sort, usually you find a solution by inspection or you don't find one at all.
The second one can be solved numerically considering, as you did, the function $$f(x)=3^{x+2} + 2^x - 5$$ By inspection, there is a root between $x=-1$ and $x=0$ (since $f(-1)=-\frac{3}{2}$ and $f(0)=5$). For the computation of this root, Newton method lookd to be a good and simple solution. Starting from an initial guess $x_0$, it will be updated according to the scheme $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, let us start iterating in the middle of the interval, $x_0=-\frac{1}{2}$. The successive iterates will be $-0.645718$, $-0.658278$, $-0.658362$ which is the solution for six significant figures.
By the way, the equation is not very well conditionned (since you looked at the graph, you know it). For solving the equation, you could consider instead $$g(x)=(x+2) \log (3)-\log \left(5-2^x\right)$$ which is very linear for $x <2$. If you expand $g(x)$ as a Taylor series built at $x=0$, you find an approximation of the solution $$x \approx -\frac{4 (2 \log (3)-\log (4))}{\log (2)+4 \log (3)} =-0.637574$$
If we consider the first equation $$f(x)=\log_{6}(1-x) + \log(x^{2}-9) - 2$$ a look at the graph shows a solution close to $x=-4$. Starting from there Newton iterations, the successive iterates are : $-3.32709$, $-3.46317$, $-3.49133$, $-3.49211$ which is the solution for six significant figures.