folve the equation $$ 1+\log (2 x-9)=\log (3 x+5)-\log 2 $$. 27. COMMERCIAI ARITHMETIC (II)

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Solve the equation $$ 1+\log (2x-9)=\log (3x+5)-\log 2 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$1+\log_{10}{\left(2x-9ight)}=\log_{10}{\left(3x+5ight)}-\log_{10}{\left(2ight)}$$
- step1: Find the domain:
  $$1+\log_{10}{\left(2x-9ight)}=\log_{10}{\left(3x+5ight)}-\log_{10}{\left(2ight)},x>\frac{9}{2}$$
- step2: Transform the expression:
  $$1+\log_{10}{\left(2x-9ight)}=\log_{10}{\left(\frac{3x+5}{2}ight)}$$
- step3: Rewrite in terms of common logarithms:
  $$\log_{10}{\left(10\left(2x-9ight)ight)}=\log_{10}{\left(\frac{3x+5}{2}ight)}$$
- step4: Set the arguments equal:
  $$10\left(2x-9ight)=\frac{3x+5}{2}$$
- step5: Cross multiply:
  $$10\left(2x-9ight)	imes 2=3x+5$$
- step6: Simplify the equation:
  $$20\left(2x-9ight)=3x+5$$
- step7: Expand the expression:
  $$40x-180=3x+5$$
- step8: Move the variable to the left side:
  $$40x-180-3x=5$$
- step9: Subtract the terms:
  $$37x-180=5$$
- step10: Move the constant to the right side:
  $$37x=5+180$$
- step11: Add the numbers:
  $$37x=185$$
- step12: Divide both sides:
  $$\frac{37x}{37}=\frac{185}{37}$$
- step13: Divide the numbers:
  $$x=5$$
- step14: Check if the solution is in the defined range:
  $$x=5,x>\frac{9}{2}$$
- step15: Find the intersection:
  $$x=5$$
The solution to the equation $$1+\log (2x-9)=\log (3x+5)-\log 2$$ is $$x=5$$.
$$x=5$$

folve the equation .
27. COMMERCIAI ARITHMETIC (II)

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folve the equation . 27. COMMERCIAI ARITHMETIC (II)