solve 7^(x+3)=e^x???

7 ^ (x + 3) = e^x

Take the natural log of both sides:

(x + 3) * ln 7 = x

x * ln 7 + 3 * ln 7 = x

3 * ln 7 = x - x * ln 7

3 * ln 7 = x * (1 - ln 7)

x = 3 * ln 7 / [ 1 - ln 7 ]

Hope that helps.

Take the ln of both sides to get

ln(7^(x+3)) = ln(e^x)

On the left you will just be left with x, because you are taking the ln of e. On the right you can bring the power to the front of the ln to get;

(x+3)ln7 = x

multiply through with ln 7

xln7 + 3ln7 = x

xln7 - x = -3ln7

x(ln7 - 1) = -3ln7

x = (-3ln7) / (ln7 - 1)


btw, this answer above is the same as the ones below.

7^(x + 3) = e^x
ln[7^(x + 3)] = x
(x + 3)ln(7) = x
xln(7) + 3ln(7) = x
xln(7) - x = -3ln(7)
x[ln(7) - 1] = -3ln(7)
x = -3ln(7)/[ln(7) - 1]

(x + 3) log 7 = x ln e

(x + 3) log 7 = x

x log 7 + 3 log 7 = x

3 log 7 = x (1 - log 7)

x = 3 log 7 / (1 - log 7)

7^(x+3) = e^x
(x+3)ln(7) = x
x(1-ln(7)) = 3ln(7)
x = 3ln(7)/(1-ln(7))

Answer: x = 3ln(7)/(1-ln(7)) .... approx. -6.17155 decimal

Yeah i agree, Billy your still grounded!

No. Do your own math homework Billy or your still grounded this weekend.