Step 1

Given: \(\log(x+3)+\log x =1\)

Step 2

Explanation:

\(\log(x+3)+\log x =1\)

\(\Rightarrow \log_{10}(x(x+3))=1\ \ \ [\because \log(m)+\log(n)=\log(mn)]\)

\(\Rightarrow x^{2}+3x=10^{1}\ \ \ [\because \log_{a}(m)=c\ then\ m = a^{c}]\)

\(\Rightarrow x^{2}+3x-10=0\)

\(\Rightarrow x^{2}+5x-2x-10=0\)

\(\Rightarrow x(x+5)-2(x+5)=0\)

\(\Rightarrow (x+5)(x-2)=0\)

\(\Rightarrow x=-5, 2\)

\(\log\) value can't be negative

So, x=2

Given: \(\log(x+3)+\log x =1\)

Step 2

Explanation:

\(\log(x+3)+\log x =1\)

\(\Rightarrow \log_{10}(x(x+3))=1\ \ \ [\because \log(m)+\log(n)=\log(mn)]\)

\(\Rightarrow x^{2}+3x=10^{1}\ \ \ [\because \log_{a}(m)=c\ then\ m = a^{c}]\)

\(\Rightarrow x^{2}+3x-10=0\)

\(\Rightarrow x^{2}+5x-2x-10=0\)

\(\Rightarrow x(x+5)-2(x+5)=0\)

\(\Rightarrow (x+5)(x-2)=0\)

\(\Rightarrow x=-5, 2\)

\(\log\) value can't be negative

So, x=2