At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
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].
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
