How to Solve a Log Equation with Different Bases?

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The discussion focuses on solving the logarithmic equation log16(3x-1) = log4(3x) + log4(0.5). The key insight is that log16(x) can be expressed in terms of log4(x) by using the relationship log16(x) = (1/2)log4(x), due to the fact that 16 is 4 squared. This allows the equation to be rewritten with a common base, facilitating the application of logarithmic laws to solve for x.

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Solve the log equation

log_16 (3x-1)=log_4 (3x) + log_4 (0.5)

Can someone get me started?

The left side has base 16 and the right side base 4.

How is this done when two different bases are involved?
 
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By definition of logarithm, $\log_{16}x$ is the number $a$ such that $16^a=x$. Note that $16^a=(4^2)^a=4^{2a}$. Therefore, if $\log_{16}x=a$, then $\log_4x=2a=2\log_{16}x$.
 
Evgeny.Makarov said:
By definition of logarithm, $\log_{16}x$ is the number $a$ such that $16^a=x$. Note that $16^a=(4^2)^a=4^{2a}$. Therefore, if $\log_{16}x=a$, then $\log_4x=2a=2\log_{16}x$.

I don't get it.
 
The quality of information you'll get as help is proportional to how much you reveal about what you understand, where the difficulty is for you and why.
 
RTCNTC said:
Solve the log equation

log_16 (3x-1)=log_4 (3x) + log_4 (0.5)

Can someone get me started?

The left side has base 16 and the right side base 4.

How is this done when two different bases are involved?
Do you know what "log_16" means? In particular, do you know that, for any positive x, if y= log_16(x) then x= 16^y? Of course 16= 4^2 so x= (4^2)^y= 4^(2y).

Now, take the logarithm base 4: since x= 4^(2y), 2y= log_4(x) so y= (1/2) log_4(x).

You can replace "log_16(3x- 1)" with "(1/2)log_4(3x- 1)" and have the same base:
(1/2)log_4(3x-1)= log_4(3x)+ log_4(0.5).

Now use the laws of logarithms to solve for x.
 
Thank you everyone.
 

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