How do you solve #7^(2x-3) - 4 = 14#?

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Answer 1 · 2016-04-06T03:50:49

You must work with the fact that #a^n = m -> loga^n = logm#

#7^(2x - 3) = 18#

#log7^(2x - 3) = log18#

Simplify using the rule #loga^n = nloga#

#(2x - 3)log7 = log18#

#2xlog7 - 3log7 = log18#

#2xlog7 = log18 + 3log7#

Use the rule #log_an + log_am = log_a(n xx m)#

#x(2log7) = log(18 xx 343)#

#x = log6174/log49#

#x = 2.24#

Make sure to ask your teacher whether or not they require your answer in logarithmic (exact) form.

Hopefully this helps!