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?
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:
$\displaystyle x\approx209.6041493$
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.