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

Question

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

1 Answer

Impact of this question

1278 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-10T21:47:46

The value of $x = 1 + \frac{log 3}{log 4}$ $x=1+log3/log4$

Explanation:

Taking logarithms in both sides we get

${log 4}^{x - 1} = log 3 \Rightarrow (x - 1) log 4 = log 3 \Rightarrow x - 1 = \frac{log 3}{log 4} \Rightarrow x = 1 + \frac{log 3}{log 4}$ $log4^(x-1)=log3=>(x-1)log4=log3=>x-1=log3/log4=>x=1+log3/log4$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 4x−1=34^(x-1)=3?

1 Answer

Explanation:

Related questions

Impact of this question

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