solve for x:
2(5)^x = 3^(x+1)
2(5)^x = 3^(x+1)
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like the other person said, take the ln (or log) of both sides
let's just use log, looks prettier :D
log[2(5)^x] = log[3^(x+1)]
log2 + log(5^x) = (x+1)log3 <----- law of logs: sum of logs and exponent law of logs
log2 + xlog5 = xlog3 + log3 <----- expansion and exponent log law
xlog5 - xlog3 = log3 - log2 <----- rearranging the equation
x(log5 - log3) = log3 - log2 <----- factoring out and singling "x"
x = (log3 - log2)/(log5 - log3) <----- divide both sides by (log5 - log3) to isolate x
and there is your answer :)
let's just use log, looks prettier :D
log[2(5)^x] = log[3^(x+1)]
log2 + log(5^x) = (x+1)log3 <----- law of logs: sum of logs and exponent law of logs
log2 + xlog5 = xlog3 + log3 <----- expansion and exponent log law
xlog5 - xlog3 = log3 - log2 <----- rearranging the equation
x(log5 - log3) = log3 - log2 <----- factoring out and singling "x"
x = (log3 - log2)/(log5 - log3) <----- divide both sides by (log5 - log3) to isolate x
and there is your answer :)
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take log both sides