$$ \begin{array}{ll} ext { a } \log _{10}\left(x^{2}-2 x+8 ight)=2 \log _{10} x & ext { b } \log _{10}(5 x)-\log _{10}(3-2 x)=1 \ ext { c } 3 \log _{10}(x-1)=\log _{10} 8 & ext { d } \log _{10}(20 x)-\log _{10}(x-8)=2 \ 2 \log _{10} 5+\log _{10}(x+1)=1+\log _{10}(2 x+7) & \ 1+2 \log _{10}(x+1)=\log _{10}(2 x+1)+\log _{10}(5 x+8) & \end{array} $$

Question

$$ \begin{array}{ll}	ext { a } \log _{10}\left(x^{2}-2 x+8ight)=2 \log _{10} x & 	ext { b } \log _{10}(5 x)-\log _{10}(3-2 x)=1 \ 	ext { c } 3 \log _{10}(x-1)=\log _{10} 8 & 	ext { d } \log _{10}(20 x)-\log _{10}(x-8)=2 \ 2 \log _{10} 5+\log _{10}(x+1)=1+\log _{10}(2 x+7) & \ 1+2 \log _{10}(x+1)=\log _{10}(2 x+1)+\log _{10}(5 x+8) & \end{array} $$

UpStudy AI · Accepted Answer

**a)**  
We start with  
$$
\log_{10}(x^2-2x+8)=2\log_{10}x.
$$  
Using the logarithm power rule, we write  
$$
2\log_{10}x=\log_{10}(x^2).
$$  
Thus, under the assumption that the arguments are positive, we have  
$$
\log_{10}(x^2-2x+8)=\log_{10}(x^2) \quad\Longrightarrow\quad x^2-2x+8=x^2.
$$  
Subtracting $$x^2$$ from both sides gives  
$$
-2x+8=0.
$$
Solving for $$x$$, we find  
$$
x=4.
$$  
The domain requires $$x>0$$ (since $$\log_{10}x$$ is defined only for positive $$x$$), and $$x=4$$ meets this condition.

---

**b)**  
We have  
$$
\log_{10}(5x)-\log_{10}(3-2x)=1.
$$  
Using the quotient rule for logarithms, we combine the two logs:  
$$
\log_{10}\left(\frac{5x}{3-2x}\right)=1.
$$  
Writing $$1$$ as $$\log_{10}10$$, we equate the arguments:  
$$
\frac{5x}{3-2x}=10.
$$  
Solve for $$x$$ by cross–multiplying:  
$$
5x=10(3-2x)=30-20x.
$$  
Then,  
$$
5x+20x=30 \quad\Longrightarrow\quad 25x=30, 
$$
so  
$$
x=\frac{30}{25}=\frac{6}{5}.
$$  
The domain requires $$5x>0$$ (i.e. $$x>0$$) and $$3-2x>0$$ (i.e. $$x<\frac{3}{2}$$). Since $$\frac{6}{5}=1.2$$ falls in this range, it is valid.

---

**c)**  
The equation is  
$$
3\log_{10}(x-1)=\log_{10}8.
$$  
Writing the left side as a single logarithm using the power rule:  
$$
\log_{10}((x-1)^3)=\log_{10}8.
$$  
Thus,  
$$
(x-1)^3=8.
$$  
Taking the cube root of both sides gives  
$$
x-1=2,
$$
so  
$$
x=3.
$$  
The domain requires $$x-1>0$$ (i.e. $$x>1$$), which is satisfied by $$x=3$$.

---

**d)**  
We consider  
$$
\log_{10}(20x)-\log_{10}(x-8)=2.
$$  
Using the quotient rule:  
$$
\log_{10}\left(\frac{20x}{x-8}\right)=2.
$$  
Express $$2$$ as $$\log_{10}100$$ (since $$10^2=100$$), so the equation becomes  
$$
\log_{10}\left(\frac{20x}{x-8}\right)=\log_{10}100.
$$
Thus,  
$$
\frac{20x}{x-8}=100.
$$  
Cross–multiply:  
$$
20x=100(x-8)=100x-800.
$$  
Subtract $$100x$$ from both sides:  
$$
20x-100x=-800 \quad\Longrightarrow\quad -80x=-800,
$$
so  
$$
x=10.
$$  
The domain requires $$20x>0$$ (i.e. $$x>0$$) and $$x-8>0$$ (i.e. $$x>8$$), so $$x=10$$ is acceptable.

---

**e)**  
The equation is  
$$
2\log_{10}5+\log_{10}(x+1)=1+\log_{10}(2x+7).
$$  
First, write $$2\log_{10}5$$ as $$\log_{10}(5^2)=\log_{10}25$$, and represent $$1$$ as $$\log_{10}10$$. Then the equation becomes  
$$
\log_{10}25+\log_{10}(x+1)=\log_{10}10+\log_{10}(2x+7).
$$  
Using the product rule on both sides, we have  
$$
\log_{10}\bigl[25(x+1)\bigr]=\log_{10}\bigl[10(2x+7)\bigr].
$$
Thus,  
$$
25(x+1)=10(2x+7).
$$
Expanding both sides:  
$$
25x+25=20x+70.
$$  
Subtract $$20x$$ from both sides:  
$$
5x+25=70,
$$  
then subtract $$25$$:  
$$
5x=45,
$$  
so  
$$
x=9.
$$  
The domain requires $$x+1>0$$ (i.e. $$x>-1$$) and $$2x+7>0$$ (i.e. $$x > -\frac{7}{2}$$), so $$x=9$$ is valid.

---

**f)**  
We are given  
$$
1+2\log_{10}(x+1)=\log_{10}(2x+1)+\log_{10}(5x+8).
$$  
Express $$1$$ as $$\log_{10}10$$ and combine $$2\log_{10}(x+1)$$ into $$\log_{10}((x+1)^2)$$. Then the left side becomes  
$$
\log_{10}\bigl[10(x+1)^2\bigr].
$$
Using the product rule on the right side, we combine the logarithms:  
$$
\log_{10}\bigl[(2x+1)(5x+8)\bigr].
$$
Hence, we have  
$$
\log_{10}\bigl[10(x+1)^2\bigr]=\log_{10}\bigl[(2x+1)(5x+8)\bigr].
$$
Thus,  
$$
10(x+1)^2=(2x+1)(5x+8).
$$
Expand both sides. On the left:  
$$
10(x^2+2x+1)=10x^2+20x+10.
$$
On the right, expand:  
$$
(2x+1)(5x+8)=10x^2+16x+5x+8=10x^2+21x+8.
$$
Set the expressions equal:  
$$
10x^2+20x+10=10x^2+21x+8.
$$
Cancel $$10x^2$$ from both sides:  
$$
20x+10=21x+8.
$$
Subtract $$20x$$ from both sides:  
$$
10= x+8,
$$
then subtract $$8$$:  
$$
x=2.
$$  
The domain requires $$x+1>0$$ (i.e. $$x>-1$$), $$2x+1>0$$ (i.e. $$x>-\frac{1}{2}$$), and $$5x+8>0$$ (i.e. $$x>-\frac{8}{5}$$), so $$x=2$$ is acceptable.

---

**Final Answers:**  
a) $$x=4$$  
b) $$x=\frac{6}{5}$$  
c) $$x=3$$  
d) $$x=10$$  
e) $$x=9$$  
f) $$x=2$$
The solutions are: a) $$x=4$$ b) $$x=\frac{6}{5}$$ c) $$x=3$$ d) $$x=10$$ e) $$x=9$$ f) $$x=2$$

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