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Solve the equation `64.9^(x)-84(12^(x))+27(16^(x))=0`

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Here `9xx16=(12)^(2)`
Then, we divide its bothsides by`12^(x)` and obtain
`implies64.(3/4)^(x)-84+27.(4/3)^(x)=0`…..i
Let `(3/4)^(x)=t`, the eq. (i) reduce in te form
`64t^(2)-84t+27=0`
`:.t_(1)=3/4` and `t_(2)=9/16`
then`(3/4)^(x)=(3/4)^(1)` and `(3/4)^(x)=(3/4)^(2)`
`:.x_(1)=1` and `x_(2)=2`
Hence, roots of the original equation are `x_(1)=1` and `x_(2)=2`

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