Answer :
To solve the equation [tex]\(\log_{10}(7x-3) + 2\log_{10}5 = 2 + \log_{10}(x+3)\)[/tex], we need to follow a step-by-step process, using the properties of logarithms.
### Step 1: Simplify the logarithmic terms
We have:
[tex]\[ \log_{10}(7x-3) + 2\log_{10}5 \][/tex]
Using the logarithmic identity [tex]\(a \log_b c = \log_b(c^a)\)[/tex], we can rewrite [tex]\(2\log_{10}5\)[/tex] as:
[tex]\[ 2\log_{10}5 = \log_{10}(5^2) = \log_{10}25 \][/tex]
So, the left-hand side of the equation becomes:
[tex]\[ \log_{10}(7x-3) + \log_{10}25 \][/tex]
### Step 2: Use the product property of logarithms
Next, apply the product rule for logarithms, which states [tex]\(\log_b(a) + \log_b(c) = \log_b(ac)\)[/tex]:
[tex]\[ \log_{10}(7x-3) + \log_{10}25 = \log_{10}((7x-3) \cdot 25) \][/tex]
So, the equation now looks like this:
[tex]\[ \log_{10}((7x-3) \cdot 25) = 2 + \log_{10}(x+3) \][/tex]
### Step 3: Isolate the logarithmic terms
To proceed, move [tex]\(\log_{10}(x+3)\)[/tex] to the left side:
[tex]\[ \log_{10}((7x-3) \cdot 25) - \log_{10}(x+3) = 2 \][/tex]
Using the quotient property of logarithms, which states [tex]\(\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\)[/tex]\):
[tex]\[ \log_{10}\left(\frac{(7x-3) \cdot 25}{x+3}\right) = 2 \][/tex]
### Step 4: Convert the logarithmic equation to an exponential equation
Recall the base-10 logarithm definition: if [tex]\(\log_{10}(A) = B\)[/tex], then [tex]\(A = 10^B\)[/tex].
Here:
[tex]\[ \log_{10}\left(\frac{(7x-3) \cdot 25}{x+3}\right) = 2 \][/tex]
This implies:
[tex]\[ \frac{(7x-3) \cdot 25}{x+3} = 10^2 = 100 \][/tex]
### Step 5: Solve the resulting rational equation
We now have:
[tex]\[ \frac{(7x-3) \cdot 25}{x+3} = 100 \][/tex]
To clear the fraction, multiply both sides by [tex]\(x+3\)[/tex]:
[tex]\[ (7x-3) \cdot 25 = 100(x+3) \][/tex]
Expand and simplify:
[tex]\[ 175x - 75 = 100x + 300 \][/tex]
Collect all [tex]\(x\)[/tex] terms on one side and constant terms on the other:
[tex]\[ 175x - 100x = 300 + 75 \][/tex]
[tex]\[ 75x = 375 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{375}{75} = 5 \][/tex]
### Step 6: Check the solution
We need to verify that [tex]\(x = 5\)[/tex] is valid by plugging it back into the original equation.
For [tex]\(x = 5\)[/tex]:
[tex]\[ \log_{10}(7(5)-3) + 2\log_{10}5 = 2 + \log_{10}(5+3) \][/tex]
Evaluate each term:
[tex]\[ \log_{10}(32) + 2\log_{10}5 = 2 + \log_{10}(8) \][/tex]
Since [tex]\(2\log_{10}5 = \log_{10}(25)\)[/tex], the left side becomes:
[tex]\[ \log_{10}(32) + \log_{10}(25) = \log_{10}(32 \cdot 25) = \log_{10}(800) \][/tex]
The right side is:
[tex]\[ 2 + \log_{10}8 = 2 + \log_{10}8 \][/tex]
Since:
[tex]\[ 2 = \log_{10}(100) \][/tex]
[tex]\[ \log_{10}(100 \cdot 8) = \log_{10}(800) \][/tex]
Both sides match, confirming [tex]\(x = 5\)[/tex] is indeed a solution.
### Conclusion
The solution to the equation [tex]\(\log_{10}(7x-3) + 2\log_{10}5 = 2 + \log_{10}(x+3)\)[/tex] is [tex]\(\boxed{x = 5}\)[/tex].
### Step 1: Simplify the logarithmic terms
We have:
[tex]\[ \log_{10}(7x-3) + 2\log_{10}5 \][/tex]
Using the logarithmic identity [tex]\(a \log_b c = \log_b(c^a)\)[/tex], we can rewrite [tex]\(2\log_{10}5\)[/tex] as:
[tex]\[ 2\log_{10}5 = \log_{10}(5^2) = \log_{10}25 \][/tex]
So, the left-hand side of the equation becomes:
[tex]\[ \log_{10}(7x-3) + \log_{10}25 \][/tex]
### Step 2: Use the product property of logarithms
Next, apply the product rule for logarithms, which states [tex]\(\log_b(a) + \log_b(c) = \log_b(ac)\)[/tex]:
[tex]\[ \log_{10}(7x-3) + \log_{10}25 = \log_{10}((7x-3) \cdot 25) \][/tex]
So, the equation now looks like this:
[tex]\[ \log_{10}((7x-3) \cdot 25) = 2 + \log_{10}(x+3) \][/tex]
### Step 3: Isolate the logarithmic terms
To proceed, move [tex]\(\log_{10}(x+3)\)[/tex] to the left side:
[tex]\[ \log_{10}((7x-3) \cdot 25) - \log_{10}(x+3) = 2 \][/tex]
Using the quotient property of logarithms, which states [tex]\(\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\)[/tex]\):
[tex]\[ \log_{10}\left(\frac{(7x-3) \cdot 25}{x+3}\right) = 2 \][/tex]
### Step 4: Convert the logarithmic equation to an exponential equation
Recall the base-10 logarithm definition: if [tex]\(\log_{10}(A) = B\)[/tex], then [tex]\(A = 10^B\)[/tex].
Here:
[tex]\[ \log_{10}\left(\frac{(7x-3) \cdot 25}{x+3}\right) = 2 \][/tex]
This implies:
[tex]\[ \frac{(7x-3) \cdot 25}{x+3} = 10^2 = 100 \][/tex]
### Step 5: Solve the resulting rational equation
We now have:
[tex]\[ \frac{(7x-3) \cdot 25}{x+3} = 100 \][/tex]
To clear the fraction, multiply both sides by [tex]\(x+3\)[/tex]:
[tex]\[ (7x-3) \cdot 25 = 100(x+3) \][/tex]
Expand and simplify:
[tex]\[ 175x - 75 = 100x + 300 \][/tex]
Collect all [tex]\(x\)[/tex] terms on one side and constant terms on the other:
[tex]\[ 175x - 100x = 300 + 75 \][/tex]
[tex]\[ 75x = 375 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{375}{75} = 5 \][/tex]
### Step 6: Check the solution
We need to verify that [tex]\(x = 5\)[/tex] is valid by plugging it back into the original equation.
For [tex]\(x = 5\)[/tex]:
[tex]\[ \log_{10}(7(5)-3) + 2\log_{10}5 = 2 + \log_{10}(5+3) \][/tex]
Evaluate each term:
[tex]\[ \log_{10}(32) + 2\log_{10}5 = 2 + \log_{10}(8) \][/tex]
Since [tex]\(2\log_{10}5 = \log_{10}(25)\)[/tex], the left side becomes:
[tex]\[ \log_{10}(32) + \log_{10}(25) = \log_{10}(32 \cdot 25) = \log_{10}(800) \][/tex]
The right side is:
[tex]\[ 2 + \log_{10}8 = 2 + \log_{10}8 \][/tex]
Since:
[tex]\[ 2 = \log_{10}(100) \][/tex]
[tex]\[ \log_{10}(100 \cdot 8) = \log_{10}(800) \][/tex]
Both sides match, confirming [tex]\(x = 5\)[/tex] is indeed a solution.
### Conclusion
The solution to the equation [tex]\(\log_{10}(7x-3) + 2\log_{10}5 = 2 + \log_{10}(x+3)\)[/tex] is [tex]\(\boxed{x = 5}\)[/tex].
