Question

13 solve the exponential equation. $125^{9x - 2} = 150$ a $-0.1069$ b $-1.8847$ c $0.3375$ d $1.0378$

Explanation:

Step1: Take log of both sides

Take the logarithm (base 10 or natural log) of both sides. Let's use natural logarithm ($\ln$). So, $\ln(125^{9x - 2})=\ln(150)$.
Using the logarithm power rule $\ln(a^b)=b\ln(a)$, we get $(9x - 2)\ln(125)=\ln(150)$.

Step2: Solve for x

First, divide both sides by $\ln(125)$: $9x - 2=\frac{\ln(150)}{\ln(125)}$.
Calculate $\frac{\ln(150)}{\ln(125)}$. $\ln(150)\approx5.0106$, $\ln(125)\approx4.8283$. So, $\frac{5.0106}{4.8283}\approx1.0378$.
Then, $9x - 2 = 1.0378$.
Add 2 to both sides: $9x=1.0378 + 2=3.0378$.
Divide by 9: $x=\frac{3.0378}{9}\approx0.3375$? Wait, no, wait. Wait, maybe I made a mistake. Wait, original equation is $125^{9x - 2}=150$. Wait, 125 is $5^3$, 150 is 1.5100. Wait, maybe I miscalculated $\ln(125)$ and $\ln(150)$. Wait, $\ln(125)=\ln(5^3)=3\ln(5)\approx3*1.6094 = 4.8282$. $\ln(150)=\ln(1.5\times100)=\ln(1.5)+\ln(100)\approx0.4055 + 4.6052 = 5.0107$. So $\frac{\ln(150)}{\ln(125)}\approx\frac{5.0107}{4.8282}\approx1.0378$. Then $9x - 2 = 1.0378$, so $9x=3.0378$, $x=\frac{3.0378}{9}\approx0.3375$? But wait, the options have C as 0.3375? Wait, no, wait the original equation: Wait, maybe I misread the exponent. Wait, the equation is $125^{9x - 2}=150$. Wait, 125 is 5^3, 150 is 1.5100. Wait, let's check the options. Wait, option C is 0.3375. Wait, but let's recalculate. Wait, maybe I used the wrong base. Let's use log base 125. Then $\log_{125}(125^{9x - 2})=\log_{125}(150)$, so $9x - 2=\log_{125}(150)$. And $\log_{125}(150)=\frac{\ln(150)}{\ln(125)}\approx1.0378$, so $9x=1.0378 + 2=3.0378$, $x=3.0378/9\approx0.3375$. Wait, but the option C is 0.3375. But wait, let's check with the equation. Let's plug x=0.3375 into 9x - 2: 90.3375 - 2=3.0375 - 2=1.0375. Then 125^1.0375. 125^1=125, 125^0.0375. Let's calculate 125^0.0375. Take natural log: 0.0375ln(125)=0.03754.8283≈0.1810. Exponentiate: e^0.1810≈1.198. So 1251.198≈149.75, which is approximately 150. So that works. Wait, but earlier I thought maybe I made a mistake, but it checks out. Wait, but the options: C is 0.3375. Wait, but let's check the options again. The options are A: -0.1069, B: -1.8847, C: 0.3375, D:1.0378. So when we solved, we got x≈0.3375, which is option C. Wait, but let's re-examine the steps.

Wait, the equation is $125^{9x - 2}=150$. Taking natural logs:

$\ln(125^{9x - 2})=\ln(150)$

$(9x - 2)\ln(125)=\ln(150)$

$9x - 2=\frac{\ln(150)}{\ln(125)}$

Calculate $\frac{\ln(150)}{\ln(125)}$:

$\ln(150) \approx 5.0106$, $\ln(125) \approx 4.8283$

$\frac{5.0106}{4.8283} \approx 1.0378$

Then:

$9x - 2 = 1.0378$

$9x = 1.0378 + 2 = 3.0378$

$x = \frac{3.0378}{9} \approx 0.3375$

Yes, that's correct. So the answer is C.

Answer:

C. 0.3375