Thread: Logarithmic Equation With Two Different Logs

1. Logarithmic Equation With Two Different Logs

Solve the log equation below. Express irrational solutions in exact form and as a decimal rounded to 3 decimal places.

log_9 (x) + 3 log_4 (x) = 14

NOTE: This question has two different logs. What is that all about?

2. use the change of base formula

3. How???

How do I use the change-of-base formula?

4. $\displaystyle \log_9{x} + 3\log_4{x} = 14$

$\displaystyle \frac{\ln{x}}{\ln{9}} + \frac{3\ln{x}}{\ln{4}} = 14$

algebra time ... isolate $\displaystyle \ln{x}$ and solve and calculate the value of x.

5. Originally Posted by magentarita
Solve the log equation below. Express irrational solutions in exact form and as a decimal rounded to 3 decimal places.

log_9 (x) + 3 log_4 (x) = 14

NOTE: This question has two different logs. What is that all about?
$\displaystyle \log_9{x+3}\log_4{x}=14$

Change to same base (10)

$\displaystyle \frac{\log{x}}{\log{9}}+\frac{3\log{x}}{\log{4}}=1 4$

$\displaystyle \frac{1}{\log{9}}\cdot\log{x}+\frac{3}{\log{4}}\cd ot\log{x}=14$

Factor out log x,

$\displaystyle \log{x}\left(\frac{1}{\log{9}}+\frac{3}{\log{4}}\r ight)=14$

$\displaystyle \log{x}=\frac{14}{\frac{1}{\log{9}}+\frac{3}{\log{ 4}}}$

Take antilog of both sides:

$\displaystyle x\approx209.6041493$

6. Thanks

I thank you both.

7. Originally Posted by magentarita
Solve the log equation below. Express irrational solutions in exact form and as a decimal rounded to 3 decimal places.

log_9 (x) + 3 log_4 (x) = 14
...
Use the base-change-formula. I'll take the natural logs (see skeeter's post):

$\displaystyle \frac{\ln{x}}{\ln{9}} + \frac{3\ln{x}}{\ln{4}} = 14$

Solve for ln(x):

$\displaystyle \ln(x) = \dfrac{14}{\dfrac1{\ln(9)}+\dfrac3{\ln(4)}} = \dfrac{14}{\dfrac{\ln(4)+3\ln(9)}{\ln(9) \cdot \ln(4)}} = \dfrac{14 \cdot \ln(9) \cdot \ln(4)}{\ln(4)+3\ln(9)}$

Now use this term as exponent to the base e to get the solution exactly:

$\displaystyle x = e^{\frac{14 \cdot \ln(9) \cdot \ln(4)}{\ln(4)+3\ln(9)}}$

For the approximate value see master's post.

8. Thank you

Thank you earboth as well.