How do you solve ${log}_{3} 3^{4 x - 1} = 15$ $Log_ 3 3^(4x-1) = 15$ ?

Question

1906 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-15T12:20:44

$x = 4$ $x=4$

Use the following property of logarithms: ${log}_{a} (b^{c}) = c \cdot {log}_{a} b$ $log_a(b^c)=c*log_a b$

Thus, the equation can be rewritten as

$(4 x - 1) ({log}_{3} 3) = 15$ $(4x-1)(log_3 3)=15$

Note that ${log}_{3} 3 = 1$ $log_3 3=1$ .

$4 x - 1 = 15$ $4x-1=15$

$x = 4$ $x=4$

How do you solve Log_ 3 3^(4x-1) = 15log334x−1=15 Log_ 3 3^(4x-1) = 15?