Question

If 64^x=1/256^y then the value of 3x+4y is​

Answer 1

Given,

`\longrightarrow\rm{64^x=(1)/(256^y)}`

We have to find out the value of 3x + 4y.

Multiplying both sides by
`\rm{256^y}` we get,

`\longrightarrow\rm{64^x\cdot256^y=1}`

Then we will write 64 and 256 as powers of 4 as,

`\rm{64=4^3}`

`256=4^4`

Then,

`\longrightarrow\rm{(4^3)^x\cdot(4^4)^y=1}`

`\longrightarrow\rm{4^(3x)\cdot4^(4y)=1\quad\quad[\because(a^m)^n=a^(mn)]}`

`\longrightarrow\rm{4^(3x+4y)=1\quad\quad[\because a^m\cdot a^n=a^(m+n)]}`

Now we will take log to the base 4.

`\longrightarrow\rm{\log_4(4^(3x+4y))=\log_4(1)}`

`\small\text{$\longrightarrow\rm{(3x+4y)\log_4(4)=\log_4(1)\quad\quad[\because\log_b(a^m)=m\cdot\log_b(a)]}$}`

We have,

`\rm{\log_4(4)=1}`

`\log_4(1)=0`

Then,

`\longrightarrow\rm{(3x+4y)(1)=0}`

`\longrightarrow\rm{\underline{\underline{3x+4y=0}}}`

Hence 0 is the answer.