Write $b$ in terms of $a$.

Question

If $a$ and $b$ are both positive and unequal, and: $$\log_ab+\log_ba^{2}=3$$

Find $b$ in terms of $a$.

Tidying up a bit, letting; $y=\log_ab$ ; and then solving the quadratic gives two solutions. $$\log_ab=1$$ and; $$\log_ab=2$$

Therefore ($a=b$) or ($a^2 = b$).

But we are told $a$ and $b$ are unequal so ($a\not=b$).

The book gives the solution ($a^2 = b$) as the answer, but surely there are no two positive and unequal values that give this solution.

What could the values for $a$ and $b$ be here?

Answer 1

$$\log_ab+\log_ba^{2}=3$$

$$\log_ab+2\log_ba=3$$

$$\log_ab+2\cdot \frac{1}{\log_ab}=3$$

Now put $x=\log_ab$ .

Then $$x+2\cdot \frac{1}{x}=3$$

For $x \neq 0$

$$x^2+2=3x$$

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

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

Therefore $x=1$ or $x=2$

Thus $\log_ab=1$ or $\log_ab=2$

That is $b=a$ or $b=a^2$

Since $a \neq b$ , $b=a^2$ .

So for any positive $a$ , we can find $b$ such that $b=a^2$.

That will satisfy the equation for any positive $a$.

How it works :

If $b=a^2$ , then $$\log_aa^2+\log_{a^2}a^{2}=2\log_aa+1=2+1$$