MHB How to Solve a Log Equation with Different Bases?

[mathdad]

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?

 

[Evgeny.Makarov]

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 $\log_4x=2a=2\log_{16}x$.

 

[mathdad]

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.

 

[Evgeny.Makarov]

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.

 

[HallsofIvy]

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.

 

[mathdad]

Thank you everyone.