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Question

Number of integral value (s) of 'x' satisfying the question
log15((secπ3)x+5)=log5(116x2)+tan13π4 is

A
0
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B
1
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C
2
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D
3
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Solution

The correct option is A 0

Solve:

log15(secπ3)x+5)=log5(116x2)+tan13π4

log51(2x+5)=log5(116x2)+1

[secπ3=2

tan(13π4)=tan(3π+π4)=1]

log5(2x+5)=log5(116x2)+log55

log5(2x+5)=log5(516x2)
[loga+logb=logab]

log5(2x+5)+log5(516x2)=c

=>log5(10x+2516x2)=0

[log51=05=1]

=10x+2516x2=1

10x+25=10x2

x2+10x+9=0

x2+x+9x+9=0

x(x+1)+9(x+1)=0

x=1 or x=9

but, 2x+5>0 and 116x2>0 beccuse
value inside log cannot be negative
or zero

2x+5>0 and 116x2>0

x>5216x2>0=>x216<0

(x4)(x+4)<0x(4,4)

So, x=1 is the only one solution

of x.

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