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Use properties of logarithms to rewrite the left hand side:
log a+log b=log(ab), so in particular,
log x+ log(x-48)=log[x(x-48)]. This gives the equation
log[(x(x-48)]=2. The equivalent equation in exponent form is 10^2=x(x-48). There are a few ways to proceed from here.
The fastest way is to recognize that x=50 is a solution, since 50(50-48)=100. Is there another solution? There is, but it's a negative number, and since the original equation contains the term log x, negative numbers are inadmissible solutions (you can't take logarithms of negative numbers, unless you're working with complex numbers).
The fast solution only works because the numbers here are "nice". A more systematic way is to expand x(x-48) to get x^2-48x, and then solve x^2-48x=100, or x^2-48x-100=0.
To solve this, either factor as follows: x^2-48x-100=(x-50)(x+2), which gives the solutions x=50, x=-2 or use the quadratic formula to get the same solutions. Again, x=50 is the only solution because log(-2) isn't defined.
`log x + log(x-48)=2`
To simplify, apply the product rule of logarithm which is `log_b a + log_b c = log_b (a*c)` .
Since the base of logarithm is not written, it indicates that its base is 10. So, we may re-write the left side as:
Then, express the logarithm to its equivalent exponent form. Note that the exponent form of `log_b a=m` is `b^m=a` .
`10^2 = x^2-48x`
Re-write the equation in quadratic form `ax^2+bx+c=0` .
Set each factor to zero and solve for x.
`x-50=0` and `x+2=0`
`x=50 ` `x=-2`
Substitute the values of x to the original equation to verify.
`x=50` , `log 50 + log (50-48)=2`
`log 50 + log 2 = 2`
`log (50*2)= 2`
`x=-2` , `log(-2)+log(-2-48)=2`
`log(-2)+log(-50)=2 ` (Invalid logarithm)
Note that in logarithm, a negative argument is not allowed. It should always be positive.
Hence, the solution to the equation `log x + log(x-48)=2` is `x=50` .
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