Logarithm Identity

Question

I am reviewing algebra 2 using some video tutorial from mathtutordvd.com.

In one of the videos, the author converts a logarithm equation into the exponent form as follows:

Which logarithm identity is used to convert the equation?

Answer 1

We are using the definition

$$\log_a b=c\iff a^c=a^{\log_a b}=b$$

but since $\log$ function is injective we can conclude directly

$$\log_a f(x)=\log_a g(x)\iff f(x)=g(x)$$

for $f(x),g(x)>0$.

Answer 2

This comes directly from the definition of a function, that for a certain input, a function produces a unique output. In other words, for a function $f$,

$$a=b \Rightarrow f(a)=f(b)$$

By definition, exponents are functions, that is, any input results in a unique output.

So, the implication made in the steps shown, albeit useful here, has nothing to do with the fact that they are logarithms of some values, but rather that for any $a,b$,

$$a=b \Rightarrow 6^a=6^b$$

Answer 3

It is perhaps easier to visualise in terms of natural logarithms. For example:

$\ln(2x-3) = \ln(4) \rightarrow \exp(\ln(2x-3)) = exp(\ln 4) $

$\rightarrow e^{\ln(2x-3)} = e^{\ln(4)} $

Just for fun and interest, we could use a little known "rule" that

$a^{\log b} = b^{\log a}$ where $a,b \neq 0,1$.

Start with $2x-3 = 4$

$\rightarrow (2x-3)^{log_6 7} = 4^{log_6 7}$

This is still valid because I have only raised both sides to the same power.

Using the "rule" $\rightarrow 7^{log_6 (2x-3)} = 7^{log_6 (4)}$

Notice that the radix is 7 and not 6 but it still works.

Nothing special about 7 either. It could be any positive real number besides 0 or 1. The whole mess can undone by applying the "rule" again. All unnecessarily complicated of course but it might be of interest ;-)

The most sensible method is simply:

$\log_6(2x-3) = \log_6(4) \rightarrow (2x-3) = (4)$

For completeness, if $a = b$ then $x^a = x^b$ is perfectly valid and as someone else mentioned, has nothing to do with the properties of logarithms.