What is the solution to the logarithmic equation with different bases?

[karush]

$\tiny{KAM}$
$\log_9{(x+1)}+3\log_3{x}=14$
ok not sure as to best approach to this
assume change the base 9?

 

[skeeter]

change

$\log_9(x+1)$ to $ \dfrac{1}{2}\log_3(x+1)$

or …

$3\log_3{x}$ to $6\log_9{x}$

either way will result in a nasty 7th degree polynomial equation that will require technology to solve

 

[karush]

by w|A $x≈80.8579$

but I didn't know you could change the base like that

 

[skeeter]

change of base formula …

$\log_b{a} = \dfrac{\log_c{a}}{\log_c{b}}$

 

[skeeter]

change of base formula …

$\log_b{a} = \dfrac{\log_c{a}}{\log_c{b}}$

derivation …

let $x = \log_b{a} \implies a = b^x$

$\log_c{a} = \log_c{b^x}$

$\log_c{a} = x \cdot \log_c{b}$

$\dfrac{\log_c{a}}{\log_c{b}} = x = \log_b{a}$

 

[karush]

change

$\log_9(x+1)$ to $ \dfrac{1}{2}\log_3(x+1)$

or …

$3\log_3{x}$ to $6\log_9{x}$

either way will result in a nasty 7th degree polynomial equation that will require technology to solve

where did the $\dfrac{1}{2}$ come from

 

[skeeter]

$\log_9(x+1) = \dfrac{\log_3(x+1)}{\log_3{9}} = \dfrac{\log_3(x+1)}{2} = \dfrac{1}{2} \log_3(x+1)$

 

[karush]

$\log_9(x+1) = \dfrac{\log_3(x+1)}{\log_3{9}} = \dfrac{\log_3(x+1)}{2} = \dfrac{1}{2} \log_3(x+1)$

well that's good to know

 

[HOI]

The "1/2" is due to the fact that 3 is the square root (1/2 power) of 9.