How do you solve ${log}_{3} z = 4 {log}_{z} 3$ $log_3z=4log_z3$ ?

Question

1925 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-29T16:10:47

$z = 9$ $z=9$

Rewrite everything using the change of base formula.

The change of base formula provides a way of rewriting a logarithm in terms of another base, like follows:

${log}_{a} b = \frac{{log}_{c} b}{{log}_{c} a}$ $log_ab=log_cb/log_ca$

In this case, the new base I will choose is $e$ $e$ , so we will use the natural logarithm.

${log}_{3} z = 4 {log}_{z} 3$ $log_3z=4log_z3$

$\Rightarrow \frac{ln z}{ln 3} = \frac{4 ln 3}{ln z}$ $=>lnz/ln3=(4ln3)/lnz$

Cross multiply.

$\Rightarrow {(ln z)}^{2} = 4 {(ln 3)}^{2}$ $=>(lnz)^2=4(ln3)^2$

Take the square root of both sides.

$\Rightarrow ln z = 2 ln 3$ $=>lnz=2ln3$

We can modify the right hand side using the rule: $b \cdot ln a = ln (a^{b})$ $b*lna=ln(a^b)$

$\Rightarrow ln z = ln (3^{2}) = ln 9$ $=>lnz=ln(3^2)=ln9$

Use the fairly intuitive rule that if $ln a = ln b$ $lna=lnb$ , then $a = b$ $a=b$ .

$\Rightarrow z = 9$ $=>z=9$

How do you solve log_3z=4log_z3log3z=4logz3log_3z=4log_z3?