Pre-Cal Help: Logarithm?

Question

Im on fall break right now and my teacher gave us serveral pages of Logarithms to do over the break, naturally, after a few days off, I dont remember how to some of them. Can some one please guide me through how to do the problem;

e^x+3=pi^x

Thanks in advance!

Answer 1

ln(e^x) + ln3 = ln(π^x)
x+ln3=xlnπ
ln3=xlnπ-x
ln3=x(lnπ-1)
x=ln3/(lnπ-1)

Answer 2

If it is e^(x+3) = pi^x, then you can take the natural log of both sides to get:


x + 3 = x log(pi)

Subtract x from both sides and you get:

3 = x(log(pi)-1)
x = 3/(log(pi)-1) = 20.7282689568

If the question is:

(e^x) + 3 = pi^x,

you are going to have to use numerical methods so solve this. We know for x=0, e^x + 3 = 4> 1 = pi^x.

On the other hand, for x>=3, e^x + 3 < pi^x.

So we know there is a solution between 0 and 3.

Finding it will require some binary searching method or Newton's method. There will be only one solution

Answer 3

Is it e^x + 3 or e^(x + 3) ?

Answer 4

ok, at the start, you like the addition property for logarithms, it relatively is: logb(xy) = logb(x) + logb(y). So, on your case, you may placed those mutually to be: log4[5(x-5)]=3 What this implies is that 4 raised to the third potential equals the contents of the parentheses, so: 4^3 = 5(x-5) sixty 4 = 5x - 25 89 = 5x x = 89/5 or 17.8 that's it!

Answer 5

Perosnally it has also been a while since ive done them but i believe you end up with something along th elines of:

X= [log(e^x + 3)] / logpi

Answer 6

Hello

So we have e^(x+3)=pi^x.

Lets multiply both sides by natural log:
ln (e^x = ln(pi^x)

Thus we have:
x +3 = ln (pi^x)
x = ln (pi^x) - 3

Hope this helps

Answer 7

e^(x+3) = pi^x <-- I assume you meant this. If so, then
ln[e^(x+3)] = ln[pi^x]
x+3 = xln(pi)
x-xln(pi) = -3
x(1-ln(pi)) = -3
x = 3/(ln(pi) -1)