Question

If log 256 base 4+ log 81 base 3- log X base 2=0 then X= ?

Answer 1

Given :-

$\sf\ \ log_4^{256} + log_3^{81}-log_2^x=0$

To Find :-

We have to find the value x

Solution:-

Now ,

$\sf\ \ log_a^b= x \\ \\ :\implies\ a=b^x$

Now similarly ,

$:\implies\sf\ log_4^{256}= a\\ \\ \\ :\implies\sf\ 256=4^a\\ \\ \\ :\implies\sf\ (4)^4=4^a\\ \\ \\ :\implies\underline{\boxed{\sf\ a=log_4^{256}=4}}$

$:\implies\sf\ log_3^{81}= b\\ \\ \\ :\implies\sf\ 81=3^a\\ \\ \\ :\implies\sf\ (3)^4=3^b\\ \\ \\ :\implies\underline{\boxed{\sf\ b=log_3^{81}=4}}$

$:\implies\sf\ log_2^{x}= c\\ \\ \\ :\implies\sf\ x=2^c\\ \\ \\ :\implies\underline{\boxed{\sf\ x=log_2^{x}=2^c}}$

$\underline{\bigstar{\red{\sf\ x=2^c}}}$

Now ,

$\dashrightarrow\sf \ log_4^{256} + log_3^{81}-log_2^x=0\\ \\ \\ \dashrightarrow\sf\ 4+4-2^c=0\\ \\ \\ \dashrightarrow\sf\ 8-2^c=0\\ \\ \\ \dashrightarrow\sf\ 8=2^c\\ \\ \\ \dashrightarrow\sf\ (2)^3=2^c\\ \\ \\ \dashrightarrow\underline{\boxed{\sf\ c=3}}$

• we have to find the value of x

$\implies\sf\ \ x= 2^c\\ \\ \\ :\implies\sf\ x=(2)^3\ \ \ \ [c=3]\\ \\ \\ :\implies\underline{\boxed{\blue{\sf\ x=8}}}$

Answer 2

Step-by-step explanation:

Given :-

\sf\ \ log_4^{256} + log_3^{81}-log_2^x=0 log

4

256

+log

3

81

−log

2

x

=0

To Find :-

We have to find the value x

Solution:-

Now ,

\begin{gathered}\sf\ \ log_a^b= x \\ \\ :\implies\ a=b^x\end{gathered}

log

a

b

=x

:⟹ a=b

x

Now similarly ,

\begin{gathered}:\implies\sf\ log_4^{256}= a\\ \\ \\ :\implies\sf\ 256=4^a\\ \\ \\ :\implies\sf\ (4)^4=4^a\\ \\ \\ :\implies\underline{\boxed{\sf\ a=log_4^{256}=4}}\end{gathered}

:⟹ log

4

256

=a

:⟹ 256=4

a

:⟹ (4)

4

=4

a

:⟹

a=log

4

256

=4

\begin{gathered}:\implies\sf\ log_3^{81}= b\\ \\ \\ :\implies\sf\ 81=3^a\\ \\ \\ :\implies\sf\ (3)^4=3^b\\ \\ \\ :\implies\underline{\boxed{\sf\ b=log_3^{81}=4}}\end{gathered}

:⟹ log

3

81

=b

:⟹ 81=3

a

:⟹ (3)

4

=3

b

:⟹

b=log

3

81

=4

\begin{gathered}:\implies\sf\ log_2^{x}= c\\ \\ \\ :\implies\sf\ x=2^c\\ \\ \\ :\implies\underline{\boxed{\sf\ x=log_2^{x}=2^c}}\end{gathered}

:⟹ log

2

x

=c

:⟹ x=2

c

:⟹

x=log

2

x

=2

c

\underline{\bigstar{\red{\sf\ x=2^c}}}

★ x=2

c

Now ,

\begin{gathered}\dashrightarrow\sf \ log_4^{256} + log_3^{81}-log_2^x=0\\ \\ \\ \dashrightarrow\sf\ 4+4-2^c=0\\ \\ \\ \dashrightarrow\sf\ 8-2^c=0\\ \\ \\ \dashrightarrow\sf\ 8=2^c\\ \\ \\ \dashrightarrow\sf\ (2)^3=2^c\\ \\ \\ \dashrightarrow\underline{\boxed{\sf\ c=3}}\end{gathered}

⇢ log

4

256

+log

3

81

−log

2

x

=0

⇢ 4+4−2

c

=0

⇢ 8−2

c

=0

⇢ 8=2

c

⇢ (2)

3

=2

c

⇢

c=3

we have to find the value of x

\begin{gathered}\implies\sf\ \ x= 2^c\\ \\ \\ :\implies\sf\ x=(2)^3\ \ \ \ [c=3]\\ \\ \\ :\implies\underline{\boxed{\blue{\sf\ x=8}}}\end{gathered}

⟹ x=2

c

:⟹ x=(2)

3

[c=3]

:⟹

x=8

• thanks