I need help on this question.
9^x + 7(3^x-1) = 16
Thanks in advance.
9^x + 7(3^x-1) = 16
Thanks in advance.
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make 9^x= 3^2x and substitute 3^x = a
then equation will be a^2+7(a-1)=16
a^2+7a-7=16
a^2+7a-23=0
so a= (-7+/-root of 141/)2 so your values will be a=2.4 and a=9.43
3^x=2.4 and 3^x=9.43
apply logs on both sides for each equation u will get x=log2.4 to the base 3
and x= log 9.43 to the base 3
then equation will be a^2+7(a-1)=16
a^2+7a-7=16
a^2+7a-23=0
so a= (-7+/-root of 141/)2 so your values will be a=2.4 and a=9.43
3^x=2.4 and 3^x=9.43
apply logs on both sides for each equation u will get x=log2.4 to the base 3
and x= log 9.43 to the base 3