Logarithmic Equation With Two Different Logs

**magentarita** · August 10th 2008, 06:24 AM

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?

**skeeter** · August 10th 2008, 06:42 AM

use the change of base formula

**magentarita** · August 10th 2008, 11:23 AM

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

**skeeter** · August 10th 2008, 11:43 AM

$\log_9{x} + 3\log_4{x} = 14$

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

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

**masters** · August 10th 2008, 01:31 PM

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?

$\log_9{x+3}\log_4{x}=14$

Change to same base (10)

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

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

Factor out log x,

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

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

Take antilog of both sides:

$x\approx209.6041493$

**magentarita** · August 10th 2008, 02:54 PM

I thank you both.

**earboth** · August 10th 2008, 09:38 PM

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):

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

Solve for ln(x):

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

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

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

For the approximate value see master's post.

**magentarita** · August 11th 2008, 04:47 AM

Thank you earboth as well.

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