Find A if 5 log x - 4 log(x^2 + 1) + 5 log(x - 1) = log A.

Step-by-step solution

1. 1

   Move every coefficient inside as an exponent. The power rule $k\log M=\log\left(M^k\right)$ applies to each term:

   $5\log x=\log\left(x^5\right),\qquad 5\log(x-1)=\log\left((x-1)^5\right),\qquad -4\log\left(x^2+1\right)=\log\left(\left(x^2+1\right)^{-4}\right)$

   The coefficient always becomes an exponent on the whole argument, so $(x^2+1)^{-4}$, never $x^{-8}+1$.

2. 2

   Turn added logs into a product. By $\log M+\log N=\log(MN)$, the three terms combine into one:

   $\log\left(x^5\left(x^2+1\right)^{-4}(x-1)^5\right)=\log A$

3. 3

   Rewrite the negative exponent as a denominator. Since $M^{-4}=\dfrac{1}{M^{4}}$, the subtracted logarithm becomes a division — which is the quotient rule $\log M-\log N=\log\frac MN$ seen from the other side:

   $A=\frac{x^5(x-1)^5}{\left(x^2+1\right)^4}$

4. 4

   Cancel the logs legitimately. $\log$ is one-to-one, so $\log P=\log Q$ forces $P=Q$ whenever both are positive. That gives the boxed value of $A$ directly, with no exponentiation needed.

5. 5

   State the domain, which the condensing quietly widens. The original left side needs $x>0$ and $x-1>0$, so $x>1$. The condensed formula would happily accept e.g. $x=0.5$, where $A>0$ but $\log(x-1)$ is undefined. Always carry $x>1$ alongside the answer.

6. 6

   Check at a convenient value. At $x=2$: the left side is $5\log 2-4\log 5+5\log 1=1.50515-2.79588+0=-1.29073$. The formula gives $A=\dfrac{32\cdot 1}{625}=0.0512$ and $\log 0.0512=-1.29073$ ✓.