# How to tackle logarithms with of the form $log_{x^2}$.

I'm trying to make sense of this logarithm, any tips on how to handle the $\log _{x^2}$ part is appreciated.

Solve the equation and find the largest (real) solution. $\log_ x(3-x) = \log_{x^2}(8-3x-x^2)$.

I tried squaring the right hand side to get rid of the $x^2$ term and have the same base on both sides but the squaring gives a rather complicated polynomial which doesn't make much sense.

• I guess you are talking about $\log x^2$ and not about $\log^2 x$ – manshu May 9 '16 at 8:24
• I tried to guess your intention, and fixed it accordingly. Please make sure it fits what you actually meant, and please use LaTex for mathematical notation. – barak manos May 9 '16 at 8:26
• Hint: $log_x(y)=z\implies x^z=y\implies (x^2)^{z/2}=y\implies \log_{x^2}y=\frac z2$ – lulu May 9 '16 at 8:47

$$\log_x(3-x)=\frac{\log_x (8-3x-x^2)}{\log_x x^2}$$

$$\log_x(3-x)=\frac{\log_x (8-3x-x^2)}{2}$$

$$(3-x)^2= (8-3x-x^2)$$

$$x^2-6x+9=(8-3x-x^2)$$

$$2x^2-3x+1=0$$

$$(2x-1)(x-1)=0$$

Hence $x=0.5$.

• But the question says the largest real solution, why did you choose the smaller one? – Yuriy S May 9 '16 at 9:23
• By $\log_x a=b$, we means $a=x^b$. When $x=1$, something interesting happen. What value should $b$ takes? For our problem, we end up with $\log_1 (2)=\log_1(4)$, of which neither is well defined. – Siong Thye Goh May 9 '16 at 15:27

$$\frac{\log_x(3-x)}{ \log x}=\frac{\log_x (8-3x-x^2)}{\log_x x^2}$$

For any base $$\frac{\log_x(3-x)}{\log_x (8-3x-x^2)}=\frac{ \log _x x}{\log_x x^2}=\frac12$$

We can remove logs by exponentiation.

$$(3-x)^2= (8-3x-x^2)$$

Simplify and factorize

$$x^2-6x+9=(8-3x-x^2)$$

$$2x^2-3x+1=0$$

$$(2x-1)(x-1)=0$$

$$x=0.5,1.0$$

Choose the smaller first root.