Question

solve
(1) log10 (x + 5) = 1
(ii) log10 (x + 1) + log10 (x - 1) = log1011 + 2 log103​

Answer 1

Answer:

1) x= 5 ( log base10 10) = 1

Answer 2

$\underline{ \underline{{ \huge{ \mathtt{Q}}} \sf UESTION \: - }}$

Solve:

$\sf \: (i) \: log_{10}(x + 5) = 1$

$\sf \: (ii) \: log_{10}(x + 1) + log_{10}(x - 1) = log_{10}(11) + 2 log_{10}(3)$

$\underline{ \underline{ { \huge{ \mathtt{A}}} \sf {NSWER - }}}$

We know that,

$\blue{ \sf{ : \implies{ log_{a}(a) = 1 }}}$

$\sf \: i) \: log_{10}(x + 5) = log_{10}(10) \\ \sf \implies \: x + 5 = 10 \\ \sf \implies \: x = 10 - 5 \\ \boxed{ \red{ \sf \implies \: x = 5}}$

we know that,

$\blue{ \sf{ : \implies{ log_{a}(b) + log_{a}(c) = log_{a}(b \times c) }}} \\ \blue{ \sf{ : \implies{2 log_{a}(b) = log_{a}( {b}^{2} ) }}}$

$\sf \: ii) \: log_{10}(x + 1) (x - 1) = log_{10}(11) + log_{10}( {3}^{2} ) \\ \sf \implies \: log( {x}^{2} + 1) = log(11 \times {3}^{2} ) \\ \sf \implies \: log( {x}^{2} - 1) = log(11 \times 9) \\ \sf \implies \: {x}^{2} - 1 = 99 \\ \sf \implies {x}^{2} = 99 + 1 \\ \sf \implies \: {x}^{2} = 100 \\ \sf \implies \: x = \sqrt{100} \\ \boxed{ \red{ \sf \implies \: x = 10}}$

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