solve for X are we correct?

[acunningham19]

ab^cx+d=h


We have x=log(h-k)/a, divided by C log(b)

are we correct

 

[tkhunny]

Maybe. You didn't start out with a 'k' and your exponent is not clear. If you mean \(\displaystyle e^{cx}\), you should include parentheses in order to be better understood. e^(cx)

 

[acunningham19]

Maybe. You didn't start out with a 'k' and your exponent is not clear. If you mean \(\displaystyle e^{cx}\), you should include parentheses in order to be better understood. e^(cx)

sorry i typed it fast it should be ab^(cx+d) +K=h

 

[HallsofIvy]

sorry i typed it fast it should be ab^(cx+d) +K=h

Then hopefully the first two steps are obvious: subtract K from both sides: ab^(cx+ d)= h- K
Divide both sides by a: b^(cx+ d)= (h- K)/a

Now, the "inverse" of an exponential is the logarithm. Take the logarithm of both sides (whatever base you want).
What do you get then?

 

[acunningham19]

Then hopefully the first two steps are obvious: subtract K from both sides: ab^(cx+ d)= h- K
Divide both sides by a: b^(cx+ d)= (h- K)/a

Now, the "inverse" of an exponential is the logarithm. Take the logarithm of both sides (whatever base you want).
What do you get then?

we got

x=log(h-k/a) +d divided by C log(b)