Simplify each term<span>.</span>
Simplify <span>3log(x)</span><span> by moving </span>3<span> inside the </span>logarithm<span>.
</span><span>log(<span>x^3</span>)+2log(y−1)−5log(x)</span><span>
</span>
Simplify <span>2log(y−1)</span><span> by moving </span>2<span> inside the </span>logarithm<span>.
</span><span>log(<span>x^3</span>)+log((y−1<span>)^2</span>)−5log(x)</span><span>
</span>
Rewrite <span>(y−1<span>)^2</span></span><span> as </span><span><span>(y−1)(y−1)</span>.</span><span>
</span><span>log(<span>x^3</span>)+log((y−1)(y−1))−5log(x)</span><span>
</span>
Expand <span>(y−1)(y−1)</span><span> using the </span>FOIL<span> Method.
</span><span>log(<span>x^3</span>)+log(y(y)+y(−1)−1(y)−1(−1))−5log(x)</span><span>
</span>
Simplify each term<span>.
</span><span>log(<span>x^3</span>)+log(<span>y^2</span>−2y+1)+log(<span>x^<span>−5</span></span>)</span><span>
</span>Remove the negative exponent<span> by rewriting </span><span>x^<span>−5</span></span><span> as </span><span><span>1/<span>x^5</span></span>.</span><span>
</span><span>log(<span>x^3</span>)+log(<span>y^2</span>−2y+1)+log(<span>1/<span>x^5</span></span>)</span><span>
</span>
Combine<span> logs to get </span><span>log(<span>x^3</span>(<span>y^2</span>−2y+1))
</span><span>log(<span>x^3</span>(<span>y^2</span>−2y+1))+log(<span>1/<span>x^5</span></span>)
</span>Combine<span> logs to get </span><span>log(<span><span><span>x^3</span>(<span>y^2</span>−2y+1)/</span><span>x^5</span></span>)</span><span>
</span>log(x^3(y^2−2y+1)/x^5)
Cancel <span>x^3</span><span> in the </span>numerator<span> and </span>denominator<span>.
</span><span>log(<span><span><span>y^2</span>−2y+1/</span><span>x^2</span></span>)</span><span>
</span>Rewrite 1<span> as </span><span><span>1^2</span>.</span>
<span><span>y^2</span>−2y+<span>1^2/</span></span><span>x^2</span>
Factor<span> by </span>perfect square<span> rule.
</span><span>(y−1<span>)^2/</span></span><span>x^2</span>
Replace into larger expression<span>.
</span>
<span>log(<span><span>(y−1<span>)^2/</span></span><span>x^2</span></span>)</span>
If the negative square root is found to be one of your solutions, then that is indicative of a pair of imaginary roots (the imaginary i). According to the conjugate rule, if you have one solution that is imaginary, you will have another but with the opposite sign. For example, if a solution to a quadratic is found to be 2 - i, then its conjugate, 2 + i is also a solution. They will ALWAYS go in pairs. Same thing with radical solutions. If one solution is found to be 
then 
will also be a solution.
Answer:
Exact Form:
√5 + 5√2
__________
5
Decimal Form:
1.86142715
…
Step-by-step explanation:
N/A
Answer:
192 markers
Step-by-step explanation:
if 8 = 4% then 2=1%
so we can assume there are 200markers in the package.
8-200= 192 markers
