Question

If m^{\log_{3}2}+2^{\log_{3}m}=mlog3​2+2log3​m= 16, then what is the value of mm​

Answer 1

Step-by-step explanation:

8

3

3

y

1

+3

y

2

+3

y

3

≥(3

y

1

+y

2

+y

3

)

1/3

3

y

1

+3

y

2

+3

y

3

≥3.(3

9

)

1/3

3

y

1

+3

y

2

+3

y

3

≥81

log

3

(3

y

1

+3

y

2

+3

y

3

)≥4

m=4

Now

3

x

1

+x

2

+x

3

≥(x

1

⋅x

2

⋅x

3

)

1/3

(

3

9

)

3

>x

1

x

2

x

3

log

3

(x

1

x

2

x

3

)≤log

3

27

log

3

x

1

+log

3

x

2

+log

3

x

3

≤3

M=3

so log

2

m

3

+log

3

M

2

=log

2

4

3

+log

3

(3)

2

=6+2=8

Answer 2

Given: mlog3​2+2log3​m= 16

To Find: the value of m

Solution:

We have,

mlog32+2log3m= 16

Now the product rule of log says, xp.xq = xp+q

loga(mn) = logam + logan

2×m log (32×3m)= 16

2m log(96m)= 16

mlog(96m)= 8

8 power m = 96m (Changed the form)

8^m= 96m

Subtracting 96m from both sides

8^M-96m = 0

m= 36.

Therefore, the value of m is 36.

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