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Find the inverse of 3 + 7^x
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3+7^x=y y-3=7^x taking ln on both sides ln7^x=ln(y-3) xln7=ln(y-3) x=ln(y-3)/ln7 switch x with y and y with x y=ln(x-3)/ln7
my teacher got f-1(x)=log7(x-3) dont know how
because whynot gave you an answer you could actually use, since you have the natural log (ln) on your calculator. your teacher just said the inverse of \[7^x\] is \[\log_7(x)\]
your teacher put solved \[x=3+7^y\] for y in two steps: \[x-3=7^y\] \[\log_7(x-3)=y\]
in other words saying \[b^x=y\] is the same as saying \[x=\log_b(y)\]
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o ok makes sense thanks for the help
yw
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