You can use properties of logarithms to simplify this, but there may be multiple ways to express this.
#4log_p3 + 2log_p2 - log_q144 = 2#
Add/Subtract:
#4log_p3 + 2log_p2 - 2 = log_q144#
Change of Base Law:
#(4(ln 3))/(ln p) + (2(ln 2))/(ln p) - 2 = (ln 144)/(ln q)#
Factor:
#(2/lnp)[2ln 3 + ln 2] - 2 = (ln 144)/(ln q)#
Multiply/Divide:
#(ln 144)/{(2/lnp)[2ln 3 + ln 2] - 2} = ln q#
Raise #e# to both sides:
#q = e^[(ln 144)/{(2/lnp)[2ln 3 + ln 2] - 2}] #
#4log_p3 + 2log_p2 - log_q144 = 2#
Add/Subtract:
#4log_p3 + 2log_p2 - 2 = log_q144#
Change of Base Law:
#(4(ln 3))/(ln p) + (2(ln 2))/(ln p) - 2 = (ln 144)/(ln q)#
#[4ln 3 + 2ln 2]/(ln p) - 2 = (ln 144)/(ln q)#
Multiply by #lnplnq#, factor out #lnq#:
#lnq([4ln 3 + 2ln 2] - 2lnp) = (ln 144)(lnp)#
Divide:
#lnq = [(ln 144)(lnp)]/([4ln 3 + 2ln 2] - 2lnp)#
Raise #e# to both sides:
#q = e^{[(ln 144)(lnp)]/([4ln 3 + 2ln 2] - 2lnp)}#
If you raise the first answer to #(lnp)/(lnp)# and multiply in the #2#, they're equivalent answers.
