`2x 3^x = 7x 5^x`

First, we will divide by x.

`==> (2) 3^x = (7)5^x`

Now, we will divide by 7.

`==> (2/7) 3^x = 5^x`

Now we will apply logarithm for both sides:

`==> log (2/7)3^x = log 5^x`

Now, we know that log ab = log a...

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`2x 3^x = 7x 5^x`

First, we will divide by x.

`==> (2) 3^x = (7)5^x`

Now, we will divide by 7.

`==> (2/7) 3^x = 5^x`

Now we will apply logarithm for both sides:

`==> log (2/7)3^x = log 5^x`

Now, we know that log ab = log a * log b

`==> log (2/7) + log 3^x = log 5^x`

We know that log (a^b)= b*log a.

`==> log (2/7) + xlog 3 = x log 5`

`==> x log 5 - xlog 3 = log (2/7)`

`==> x ( log5 - log 3)= log (2/7)`

Now we know that log (a/b)= log a - log b ==> log (2/7)= log 2 - log 7

`==> x = (log 2 - log 7)/(log 5 - log3)`

``

We have to solve 2*3^x = 7*5^x

2*3^x = 7*5^x

take the log of both the sides

log 2 + x*log 3 = log 7 + x*log 5

=> x(log 3 - log 5) = log 7 - log 2

=> x = (log 7 - log 2)/(log 3 - log 5)

=> x= -2.4524

**The solution of the equation is x = -2.4524**