# SOLUTION: how do you write 5^x-7=9^-9x as a 10 logrithm finding the value of x. Thanks for your help

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 Question 536128: how do you write 5^x-7=9^-9x as a 10 logrithm finding the value of x. Thanks for your helpFound 2 solutions by fcabanski, lwsshak3:Answer by fcabanski(874)   (Show Source): You can put this solution on YOUR website!Remember that log and exponents are inverse such that: If b^y=x then log_b(x) = y. 5^(x-7)=9^(-9x) can be rewritten as: log_5(9^(-9x))=x-7 Remember log (x^y) = y*log(x). Rewrite as: (-9x)*log_5(9)=x-7 Convert log_x to log_y with log_y(z) = log_x(z)/log_x(y) We can convert from base-5 to base-10 in this case. That gives: (-9x)*(log_10(9)/log_10(5)) = x-7. Now evaluate the logs and complete the algebra. -9x*(1.36521239)=x-7 ---> -12.2869115x=x-7 (subtract x from both sides) -13.2869115x=-7 (divide both sides by -13.2869115) x=0.526834246 I provide online tutoring (\$30/hr) and personal problem solving (\$3.50-\$5.50 per problem. Contact me (check the profile "website" which is my email address. Answer by lwsshak3(6469)   (Show Source): You can put this solution on YOUR website!how do you write 5^x-7=9^-9x as a 10 logrithm finding the value of x. .. 5^(x-7)=9^-9x take log of both sides using exponent rule (x-7) log5=-9xlog9 (x-7)/9x=-log9/log5 using calculator (x-7)/9x=-1.3652 x-7=9x*-1.362=-12.2869x x+12.2869x=7 13.2869x=7 x=7/13.2869≈.5268