The expression which is equivalent to [tex]5 log_{10}x+log_{10}20-log_{10}10[/tex] is
[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex].
A logarithmic equation exists as an equation that applies the logarithm of an expression having a variable.
Product Rule Law: [tex]log_{a} (MN) = log_{a} M + log_{a} N[/tex]
Quotient Rule Law: [tex]log_{a} (M/N) = log_{a} M - log_{a} N[/tex]
Power Rule Law: [tex]log_{a}M^{n} = n log_{a} M[/tex]
Given: [tex]5 log_{10}x+log_{10}20-log_{10}10[/tex]
[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5) +log_{10}20 -log_{10}10[/tex]
apply the law of logarithm
[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*20/10)[/tex]
[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*2)[/tex]
[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(2x^5)[/tex]
Another possible equivalent expression is:
[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*20)-log_{10}10[/tex]
[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-log_{10}10[/tex]
Substitute the value of [tex]log_{10}10 = 1[/tex]
[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex]
Therefore, the correct answer
[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex].
To learn more about logarithm refer to:
https://brainly.com/question/23861954
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