+130511
I assume you want the "inverse" of this ???
y = 1.029^x + 2.25 subtract 2.25 from both sides
y - 2.25 = 1.029^x take the log of both sides
log [ y - 2.25] = log 1.029^x and we can write
log [ y - 2.25 ] = x log 1.029 divide both sides by log1.029
log [ y - 2.25 ] / log 1.029 = x "exchange" x and y
log [ x - 2.25 ] / log 1.029 = y and for y, we can write f-1(x)
So the inverse is
f-1(x) = log [ x - 2.25 ] / log 1.029
Here's the graph of both.....https://www.desmos.com/calculator/zjdmsqygfg
Notice that the graph is symmetric to y = x.....as we would expect....
+130511
I assume you want the "inverse" of this ???
y = 1.029^x + 2.25 subtract 2.25 from both sides
y - 2.25 = 1.029^x take the log of both sides
log [ y - 2.25] = log 1.029^x and we can write
log [ y - 2.25 ] = x log 1.029 divide both sides by log1.029
log [ y - 2.25 ] / log 1.029 = x "exchange" x and y
log [ x - 2.25 ] / log 1.029 = y and for y, we can write f-1(x)
So the inverse is
f-1(x) = log [ x - 2.25 ] / log 1.029
Here's the graph of both.....https://www.desmos.com/calculator/zjdmsqygfg
Notice that the graph is symmetric to y = x.....as we would expect....
