ONLY THE BRAINLIEST CAN ANSWER THIS:
SOLVE:
Answers
Step-by-step explanation:
log3(x+5x+1)+log9(x+5)2=1
Solve it .
Solution :
\sf\frac{1}{2} log_{ \sqrt{3} }( \frac{x + 1}{x + 5} ) + log_{9}( {x + 5)}^{2} = 121log3(x+5x+1)+log9(x+5)2=1
\implies \sf \: \frac{1}{2} . \frac{ log( \frac{x + 1}{x + 5} ) }{ log( \sqrt{3}) } + \frac{ log( {x + 5)}^{2} }{ log(9) } = 1 \: \: [identity : \: log_{a}(b) = \frac{ log(b) }{ log(a) } ]⟹21.log(3)log(x+5x+1)+log(9)log(x+5)2=1[identity:loga(b)=log(a)log(b)]
\implies \sf \: \frac{1}{2} . \frac{ log( \frac{x + 1}{x + 5} ) }{ \frac{1}{2} log(3) } + \frac{2 log(x + 5) }{2 log(3) } = 1⟹21.21log(3)log(x+5x+1)+2log(3)2log(x+5)=1
\implies \sf \: \frac{ log( \frac{x + 1}{x + 5} ) }{ log(3) } + \frac{ log(x + 5) }{ log(3) } = 1⟹log(3)log(x+5x+1)+log(3)log(x+5)=1
\implies \sf \: log_{3}( \frac{x + 1}{x + 5} ) + log_{3}(x + 5) = 1 \: \: [identity : \frac{ log(b) }{ log(a) } = log_{a}(b) ]⟹log3(x+5x+1)+log3(x+5)=1[identity:log(a)log(b)=loga(b)]
\implies \sf \: log_{3}( \frac{x + 1}{x + 5} .( x + 5)) = 1 \: \: [identity : log_{a}(m) + log_{a}(n) = log_{a}(mn) ]⟹log3(x+5x+1.(x+5))=1[identity:loga(m)+loga(n)=loga(mn)]
\implies \sf \: log_{3}(x + 1) = 1⟹log3(x+1)=1
\implies \sf \: x + 1 = 3 \: \:[identity : log_{a}(b) = x \implies {a}^{x} = b]⟹x+1=3[identity:loga(b)=x⟹ax=b]
\implies \sf \: x = 3 - 1⟹x=3−1
\implies\sf{x=2}⟹x=2
Therefore, the value of x is 2.
Step-by-step explanation:
Put all logarithms to one side:
3
log
5
x
−
log
5
(
5
x
)
+
log
5
25
=
3
log
5
25
can be rewritten as
log
25
log
5
=
2
log
5
log
5
=
2
log
5
x
3
−
log
5
(
5
x
)
+
2
=
3
log
5
x
3
−
log
5
5
x
=
1
log
5
(
x
3
5
x
)
=
1
x
2
5
=
5
1
x
2
=
25
x
=
±
5
The
−
5
solution is extraneous, since
log
5
(
a
x
)
, where
a
is a positive constant is only defined in the positive-x-values.
Hopefully this helps!