F Question 22 of $$ 28 \rightarrow $$ This te $$ \begin{array}{l}\text { Find } x \text { in the following equation. } \ \log _{10}(x+5)-\log _{10}(x-5)=1 \ x=\square \text { (Type a fraction or an integer. Simplify your answer.) }\end{array} $$

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Solve the equation $$ \log_{10}(x+5)-\log_{10}(x-5)=1 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\log_{10}{\left(x+5ight)}-\log_{10}{\left(x-5ight)}=1$$
- step1: Find the domain:
  $$\log_{10}{\left(x+5ight)}-\log_{10}{\left(x-5ight)}=1,x>5$$
- step2: Transform the expression:
  $$\log_{10}{\left(\frac{x+5}{x-5}ight)}=1$$
- step3: Convert the logarithm into exponential form:
  $$\frac{x+5}{x-5}=10^{1}$$
- step4: Evaluate the power:
  $$\frac{x+5}{x-5}=10$$
- step5: Cross multiply:
  $$x+5=\left(x-5ight)	imes 10$$
- step6: Simplify the equation:
  $$x+5=10\left(x-5ight)$$
- step7: Expand the expression:
  $$x+5=10x-50$$
- step8: Move the expression to the left side:
  $$x-10x=-50-5$$
- step9: Add and subtract:
  $$-9x=-50-5$$
- step10: Add and subtract:
  $$-9x=-55$$
- step11: Change the signs:
  $$9x=55$$
- step12: Divide both sides:
  $$\frac{9x}{9}=\frac{55}{9}$$
- step13: Divide the numbers:
  $$x=\frac{55}{9}$$
- step14: Check if the solution is in the defined range:
  $$x=\frac{55}{9},x>5$$
- step15: Find the intersection:
  $$x=\frac{55}{9}$$
The solution to the equation $$ \log_{10}(x+5)-\log_{10}(x-5)=1 $$ is $$ x = \frac{55}{9} $$ or $$ x = 6.\dot{1} $$.
$$ x = \frac{55}{9} $$

F Question 22 of This te

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