For questions [tex]$10-13$[/tex], solve the equation for [tex]$x$[/tex]. Round to two decimal places.

10. [tex]$\log_4 x = 3$[/tex]

11. [tex]$\ln x = -2.5$[/tex]

12. [tex]$2 \cdot \log_3 x + 4 = 1$[/tex]

13. [tex]$10 \cdot \log (2x) = 30$[/tex]



Answer :

Sure! Let's solve each of the given equations for [tex]\( x \)[/tex] step by step and round the answers to two decimal places.

### 10. [tex]\(\log_4 x = 3\)[/tex]

To solve this equation, we need to rewrite it in its exponential form:
[tex]\[ x = 4^3 \][/tex]
Now, calculate [tex]\( 4^3 \)[/tex]:
[tex]\[ 4^3 = 64 \][/tex]

So, the solution is:
[tex]\[ x = 64 \][/tex]

### 11. [tex]\(\ln x = -2.5\)[/tex]

To solve this logarithmic equation, we need to rewrite it in its exponential form using the natural logarithm property:
[tex]\[ x = e^{-2.5} \][/tex]
Calculate [tex]\( e^{-2.5} \)[/tex]:
[tex]\[ e^{-2.5} \approx 0.08 \][/tex]

So, the solution is:
[tex]\[ x \approx 0.08 \][/tex]

### 12. [tex]\(2 \cdot \log_3 x + 4 = 1\)[/tex]

First, isolate the logarithmic term:
[tex]\[ 2 \cdot \log_3 x = 1 - 4 \][/tex]
[tex]\[ 2 \cdot \log_3 x = -3 \][/tex]

Now, divide by 2:
[tex]\[ \log_3 x = -1.5 \][/tex]

Rewrite in its exponential form:
[tex]\[ x = 3^{-1.5} \][/tex]

Calculate [tex]\( 3^{-1.5} \)[/tex]:
[tex]\[ 3^{-1.5} \approx 0.19 \][/tex]

So, the solution is:
[tex]\[ x \approx 0.19 \][/tex]

### 13. [tex]\(10 \cdot \log (2x) = 30\)[/tex]

First, isolate the logarithmic term:
[tex]\[ \log (2x) = \frac{30}{10} \][/tex]
[tex]\[ \log (2x) = 3 \][/tex]

Rewrite it in its exponential form:
[tex]\[ 2x = 10^3 \][/tex]

Calculate [tex]\( 10^3 \)[/tex]:
[tex]\[ 10^3 = 1000 \][/tex]

Now, solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = 1000 \implies x = \frac{1000}{2} = 500 \][/tex]

So, the solution is:
[tex]\[ x = 500.00 \][/tex]

### Summary

The solutions to the equations are:

10. [tex]\( x = 64 \)[/tex]
11. [tex]\( x \approx 0.08 \)[/tex]
12. [tex]\( x \approx 0.19 \)[/tex]
13. [tex]\( x = 500.00 \)[/tex]