+0

# idk how to solve this I am suppose to get type close to each other

0
132
2

850e^(0.055x)=195e^(0.075x)

Guest Mar 20, 2017
Sort:

#1
0

Solve for x over the real numbers:
850 e^(0.055 x) = 195 e^(0.075 x)

850 e^(0.055 x) = 850 e^(11 x/200) and 195 e^(0.075 x) = 195 e^(3 x/40):
850 e^((11 x)/200) = 195 e^((3 x)/40)

Take the natural logarithm of both sides and use the identities log(a b) = log(a) + log(b) and log(a^b) = b log(a):
(11 x)/200 + log(850) = (3 x)/40 + log(195)

Subtract (3 x)/40 + log(850) from both sides:
-x/50 = log(195) - log(850)

Multiply both sides by -50:
Answer: | x = 50 log(850) - 50 log(195)= 73.61183954603077.......

Guest Mar 20, 2017
#2
+18607
0

850e^(0.055x)=195e^(0.075x)

$$\begin{array}{|rcll|} \hline 850\cdot e^{0.055x} &=& 195\cdot e^{0.075x} \quad & | \quad : 195 \\ \frac{850}{195} \cdot e^{0.055x} &=& e^{0.075x} \quad & | \quad \cdot e^{-0.055x} \\ \frac{850}{195} &=& e^{0.075x} \cdot e^{-0.055x} \\ \frac{850}{195} &=& e^{0.075x-0.055x} \\ \frac{850}{195} &=& e^{0.02x} \\ \frac{170}{39} &=& e^{0.02x} \quad & | \quad \text{ln of both sides } \\ \ln(\frac{170}{39}) &=& \ln(e^{0.02x}) \\ \ln(\frac{170}{39}) &=& 0.02x\cdot \ln(e) \quad & | \quad \ln(e) = 1 \\ \ln(\frac{170}{39}) &=& 0.02x \quad & | \quad : 0.02 \\ \frac{ \ln(\frac{170}{39}) } {0.02} &=& x \\ \frac{ \ln(4.35897435897) } {0.02} &=& x \\ \frac{ 1.47223679092 } {0.02} &=& x \\ 73.6118395460 &=& x \\ \hline \end{array}$$

heureka  Mar 20, 2017

### 18 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details