How do you solve for #x#? #log_b x = log_b 8^(2//3) + log_b 9^(1//2) - log_b 6#

#x = 2#

You can use the change of base law:

#log_color(red)(b) color(green)(a) = (log_10 a)/(log_10 b) = (color(green)(ln a))/(color(red)(ln b))#

Each of these then becomes:

#(ln x)/(ln b) = 2/3 (ln 8)/(ln b) + 1/2 (ln 9)/(ln b) - (ln 6)/(ln b)#

Now one can multiply through by #ln b#, so it doesn't matter what #b# was.

#ln x = 2/3 ln8 + 1/2 ln9 - ln6#

With some further log properties:

log coefficient property

• #clna = ln a^c#

log addition properties

• #ln a pm ln b = { (ln (ab), (+)),(ln (a//b), (-)) :}#

We then get:

#ln x = ln8^"2/3" + ln9^"1/2" - ln6#

#= ln((8^"2/3" cdot 9^"1/2")/(6))#

Now, use the property that

#e^(lnx) = x#

to get

#x = e^(ln((8^"2/3" cdot 9^"1/2")/(6))) = (8^"2/3" cdot 9^"1/2")/(6)#

This means we get:

#color(blue)(x) = ((8^2)^"1/3" cdot 9^"1/2")/(6) = (root(3)(64) cdot sqrt9)/(6)#

#= (4 cdot 3)/(6) = color(blue)(2)#

#"using the "color(blue)"laws of logarithms"#

#•color(white)(x)logx+logy=log(xy)#

#•color(white)(x)logx-logy=log(x/y)#

#•color(white)(x)logx^nhArrnlogx#

#•color(white)(x)logx=logyrArrx=y#

#"consider the right side"#

#rArrlog_b8^(2/3)+log_b9^(1/2)-log_b6#

#=log_b((8^(2/3)xx9^(1/2))/6)#

#=log_b((4xx3)/6)=log_b2#

#rArrlog_bx=log_b2rArrx=2#

As the everything is in logs of the same base we may work back from the logs to the 'source' values.

Taking each part separately

#log_b(x)->log_b(x) larr" No change"#

#2/3log_b(8)->log_b(8^(2/3)) larr" Changed"#

#1/2log_b(9)->log_b(9^(1/2)) larr" Changed"#

#log_b(6)->log_b(6) larr" No change"#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Putting it all together

#log_b(x)=log_b(8^(2/3))+log_b(9^(1/2)) -log_b(6)#

In logs:
Add is the consequence of multiplication of the source values.
Subtraction is the consequence of division of the source values.

So we can now write as:

#log_b(x)=log_b((8^(2/3)xx9^(1/2))/6)#

As both sides are logs then the same is true if both sides are not logs. Thus we have:

#x=(8^(2/3)xx9^(1/2))/6#

Using 'principle roots' ( the positive versions )

Note that #8^(2/3) = 8^(1/3xx2)=(root(3)(8))^2 = 2^2=4#
Note that #9^(1/2)=sqrt(9)=3#

Giving:

#x=(4xxcancel(3)^1)/(cancel(6)^2 )= 2#