Pre-Calc Question: Logarithms?

Question

Solve analytically.

5^x = 4^(x+3)

Answer 1

5^x = 4^(x+3)
log5^x = log4^(x+3)
xlog5 = (x+3)log4
xlog5 = xlog4 + 3log4
xlog5 - xlog4 = 3log4
x(log5 - log4) = 3log4
x = 3log4/(log5 - log4)
x = 18.638 ANS

teddy boy

Answer 2

a million) enable y = 8^{log2 (3)} ....... (a million) and enable log2 (3) = x, => 2^x = 3 ....... (2) Now, y = 8^x = (2^3)^x = 2^(3x) = (2^x)^3 = 3^3 = 27 [ from (2) 2^x = 3 ] 2) 2^(x + a million) = 3x ; [ is this assertion superb ? Is the RHS 3x or 3^x ?] => (2^x)*2 = 3x => (3/2)x = 2^x [ Take logarithm of the two components, the backside is meant to be 10 ] => log x + log (3/2) = x log 2 => log x + (log 3 - log 2) = x log 2 => log x + log 3 = x log 2 + log 2 = (a million + x) log 2 i think of, you haven't any longer pronounced the 2nd query wisely. Please be sure.

Answer 3

x log 5 = (x + 3) log 4
(log 5 - log 4) x = 3 log 4
x = 3 log 4 / (log 5 - log 4)
x = 3 log 4 / log (5/4)
x = 18.6 (to 1 dec. place)

Answer 4

xlog(5) = (x+3)log(4)
(x+3)/x = log(5)/log(4) = a

x+3 = ax
x(a-1) = 3
x = 3/(a-1)

Answer 5

Take the log of both sides of the equation, using the relation log(x^y) - y*log(x):

x*log(5) = (x+3)*log(4)

x*[log(5) - log(4)] = 3*log(4)

x = 3*log(4) / [log(5) - log(4)]

x = 3*log(4) / log(5/4) = 3*log(4) / log(1.25)

x = 18.64

Answer 6

18.6.....