Question 1090834

can someone help me with this?
Convert y=3*2^x to the continuous model y=Pe^rx (round r=0.0001)
Convert y=32*e^0.15x to  the standard model y=ab^x (round b=0.0001)
Thank you 
<pre>{{{matrix(1,3, y, "=", 3(2)^x)}}}
{{{matrix(1,3, y, "=", P(e^r)^x))}}} <======== Compare the above to this
As seen, {{{matrix(1,3, 2, "=", e^r)}}}
{{{matrix(1,3, r, "=", ln (2))}}} ----------- Converting to LOGARITHMIC (Natural) form
{{{matrix(1,3, r, "=", 0.693147181)}}}, so {{{highlight_green(matrix(1,5, y, "=", 3(2)^x, "becomes:", highlight(matrix(1,3, y, "=", 3e^(0.6931x)))))}}}
I take it that rounding r to .0001 (as per above instructions) means that r should be rounded to 4 decimal places. 


{{{matrix(1,3, y, "=", 32(e^0.15)^x)}}}
{{{matrix(1,3, y, "=", a(b)^x))}}} <========== Compare the above to this
As seen, {{{matrix(1,3, e^0.15, "=", b)}}}
{{{matrix(1,3, b, "=", 1.161834243)}}}, so {{{highlight_green(matrix(1,5, y, "=", 32e^(.015x), "becomes:", highlight(matrix(1,3, y, "=", 32(1.1618)^x)))))}}}
I take it that rounding b to .0001 (as per above instructions) means that r should be rounded to 4 decimal places.</pre>