${log}_{10} (x + 1) + l o g_{10} (x - 1) = l o g_{10} 11 + 2 l o g_{10} 3$
Calculate the value of x.

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Solution

${log}_{10} (x + 1) + l o g_{10} (x - 1) = l o g_{10} 11 + 2 l o g_{10} 3$
We know ${log}_{a} m + l o g_{a} n = l o g_{a} m n$ and
$n {log}_{a} m = l o g_{a} m^{n}$
Hence
${log}_{10} (x + 1) \times (x - 1) = l o g_{10} 11 \times 9$
${log}_{10} (x^{2} - 1) = l o g_{10} 99$
Comparing LHS and RHS
$x^{2} - 1 = 99$
$x^{2} = 100$
$x = 10$

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