Question

Solve 3^(2x) = 7^(x-1) −7.74293 7.74293 −1 1

Answer 1

Hello,

3^(2x)=7^(x-1)
==>2x ln 3 = (x-1)ln 7
==>x(2ln 3 -ln 7)=-ln 7
==>x=-ln 7 /(2 ln 3-ln 7)
==>x=-7,7429304889740634853601918674823...

Answer A

Answer 2

Option (a) is correct.

`x=-7.74293`

Explanation:

Given :
`\:3^(\left(2x\right))\:=\:7^(\left(x-1\right))`

We have to solve the given expression
`\:3^(\left(2x\right))\:=\:7^(\left(x-1\right))`

Consider the given expression,
`\:3^(\left(2x\right))\:=\:7^(\left(x-1\right))`

Using,
`\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)` , we have,

`\ln \left(3^(2x)\right)=\ln \left(7^(x-1)\right)`

Apply log rule,

`\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)`

We have,

`\ln \left(3^(2x)\right)=2x\ln \left(3\right),\:\space\ln \left(7^(x-1)\right)=\left(x-1\right)\ln \left(7\right)`

substitute, we get,

`2x\ln \left(3\right)=\left(x-1\right)\ln \left(7\right)`

`\mathrm{Expand\:}\left(x-1\right)\ln \left(7\right):\quad \ln \left(7\right)x-\ln \left(7\right)`

Subtract
`\ln \left(7\right)x` from both side, we have,

`2x\ln \left(3\right)-\ln \left(7\right)x=\ln \left(7\right)x-\ln \left(7\right)-\ln \left(7\right)x`

Simplify, we have,

`2x\ln \left(3\right)-\ln \left(7\right)x=-\ln \left(7\right)`

Taking x common from both term , we have,

`x\left(2\ln \left(3\right)-\ln \left(7\right)\right)`

Divide both side by
`2\ln \left(3\right)-\ln \left(7\right)` , we have,

`(\left(2\ln \left(3\right)-\ln \left(7\right)\right)x)/(2\ln \left(3\right)-\ln \left(7\right))=(-\ln \left(7\right))/(2\ln \left(3\right)-\ln \left(7\right))`

On simplify , we get,

`x=-(\ln \left(7\right))/(\ln \left((9)/(7)\right))`

Thus,
`x=-7.74293.....`

Thus, option (a) is correct.