4) `log_2(3-2x)=2 `

Rewrite as an exponential equation. (Note that by definition, `log_by=x ` if and only if `y=b^x ` )

`2^2=3-2x `

4=3-2x

`x=-1/2 `

With logarithmic equations, you must always check to make sure that you have not introduced extraneous solutions:

`log_2(3-2(-1/2))=log_2(4)=2 `

**The solution is** `x=-1/2 `

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5) `"lg"^2x+"lg"x-2=0 `

This is a quadratic equation in lg -- you can solve it using the quadratic formula, or in this case it factors:

`"lg"^2x+"lg"x-2=0 ==> ("lg"x+2)("lg"x-1)=0 `

Then lgx=-2 or lgx=1.

For lgx=1 the solution is x=0 for any legitimate base for the logarithm.

For lgx=-2, the solution is `x=b^(-2) ` where b is the base of the logarithm. (If lgx is the natural logarithm (base e) the solution is `x=e^(-2) ` , while if lgx is the common logarithm (base 10) the solution is x=.01)