Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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].
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.