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Basics of Mathematics
NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Basics of Mathematics
NDA 2017 · Mathematics · Basics of Mathematics
If x+ _ 10 (1+2^x)=x _ 10 5+ _ 10 6 then x is equal to
If
x
+
log
10
(
1
+
2
x
)
=
x
log
10
5
+
log
10
6
x+\log _{10}\left(1+2^x\right)=x \log _{10} 5+\log _{10} 6
x
+
lo
g
10
(
1
+
2
x
)
=
x
lo
g
10
5
+
lo
g
10
6
then
x
x
x
is equal to
Options
A
2
,
−
3
2,-3
2
,
−
3
B
2
2
2
only
C
1
1
1
D
3
3
3
Correct answer
C.
1
1
1
Step-by-step solution
x
+
log
10
(
1
+
2
x
)
=
x
log
10
5
+
log
10
6
⇒
x
−
x
log
10
5
=
log
10
6
−
log
10
(
1
+
2
x
)
⇒
x
(
1
−
log
10
5
)
=
log
10
6
−
log
10
(
1
+
2
x
)
⇒
x
(
log
10
10
−
log
10
5
)
=
log
10
(
6
1
+
2
x
)
⇒
x
(
log
10
(
10
5
)
)
=
log
10
(
6
1
+
2
x
)
⇒
x
log
10
2
=
log
10
(
6
1
+
2
x
)
\begin{aligned} & x+\log _{10}\left(1+2^x\right)=x \log _{10} 5+\log _{10} 6 \\ & \Rightarrow x-x \log _{10} 5=\log _{10} 6-\log _{10}\left(1+2^x\right) \\ & \Rightarrow x\left(1-\log _{10} 5\right)=\log _{10} 6-\log _{10}\left(1+2^x\right) \\ & \Rightarrow x\left(\log _{10} 10-\log _{10} 5\right)=\log _{10}\left(\frac{6}{1+2^x}\right) \\ & \Rightarrow x\left(\log _{10}\left(\frac{10}{5}\right)\right)=\log _{10}\left(\frac{6}{1+2^x}\right) \\ & \Rightarrow x \log _{10} 2=\log _{10}\left(\frac{6}{1+2^x}\right)\end{aligned}
x
+
lo
g
10
(
1
+
2
x
)
=
x
lo
g
10
5
+
lo
g
10
6
⇒
x
−
x
lo
g
10
5
=
lo
g
10
6
−
lo
g
10
(
1
+
2
x
)
⇒
x
(
1
−
lo
g
10
5
)
=
lo
g
10
6
−
lo
g
10
(
1
+
2
x
)
⇒
x
(
lo
g
10
10
−
lo
g
10
5
)
=
lo
g
10
(
1
+
2
x
6
)
⇒
x
(
lo
g
10
(
5
10
)
)
=
lo
g
10
(
1
+
2
x
6
)
⇒
x
lo
g
10
2
=
lo
g
10
(
1
+
2
x
6
)
This is possible only when
x
=
1
x=1
x
=
1
.
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