p(x) = log 6 + 5 sq
Answers
Answered by
1
Answer:
Answer should be 2ab-1
By BASE CHANGE RULE
Log5(base)4 become log5÷log4
Similarly log6÷log5
Multiply both we get
(Log6×log5)÷(log5×log4)=a×b
log6÷log4=a×b
Log(3×2)÷log4=a×b
(log3+log2)÷2log2=a×b
log3÷2log2+1/2=a×b
Log3÷2log2=ab-1/2
1/2log3÷log2=( 2ab-1)/2
By base change RULE
Log3(base)2=(2ab-1)2/2
Log3(base)2=2ab-1
Answered by
1
Solve for x log base 6 of x+ log base 6 of x-5=2
log
6
(
x
)
+
log
6
(
x
−
5
)
=
2
Simplify the left side.
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log
6
(
x
2
−
5
x
)
=
2
Rewrite
log
6
(
x
2
−
5
x
)
=
2
in exponential form using
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