Solve for x:

a) 2^4x-1 = 3^1-x

b) 8^x=2^x+1... for this one I got 1/2 as my answer. is that correct?

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You should notice that the bases are different, hence you need to use logarithms to solve the equation such that:

`2^(4x-1) = 3^(1-x) =gt ln 2^(4x-1) = ln 3^(1-x)`

You need to use the power property of logarithms such that:

`(4x-1)ln2 = (1-x)ln 3`

You should open the brackets such that:

`4x ln 2 - ln 2 = ln 3 - x ln 3`

You need to move the terms containing x to the left side such that:

`4x ln 2 + x ln 3 = ln 3 + ln 2`

You should factor out x such that:

`x(4 ln 2 + ln 3) = ln (3*2)` (product property of logarithms)

`x(ln 2^4 + ln 3) = ln 6`

`x ln 48 = ln 6 =gt x = (ln 6)/(ln 48)`

Hence, evaluating the solution to equation yields `x = (ln 6)/(ln 48).`

b) You need to write `8 = 2^3 =gt 8^x = (2^3)^x = 2^(3x)`

You need to solve the equation `2^(3x) = 2^(x+1) ` such that:

`2^(3x) = 2^(x+1) =gt 3x = x+1 =gt 2x = 1 =gt x = 1/2`

**Hence, you need to notice that your answer is good, thus the solution to the equation is `x = 1/2` .**

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