Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

Solve for [tex][tex]$x$[/tex][/tex] in the equation [tex][tex]$\log_{10}(7x - 3) + 2 \log_{10} 5 = 2 + \log_{10}(x + 3)$[/tex][/tex].

(6 marks)


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].
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.