# If x^3x = (3x)^4x, then find the value of x.Last time i wrote 3x^4x but it was (3x)^4x... Plz solve it...

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### 4 Answers

You should take logarithms both sides such that:

`ln (x^(3x)) = ln((3x)^(4x))`

Using the logarithmic identity `ln (x^y) = y ln x` yields:

`3x*(ln x) = 4x*ln (3x)`

You should convert the logarithm of product in sum of logarithms such that:

`ln (3x) = ln 3 + ln x`

`3x*(ln x) = 4x*(ln 3 + ln x)`

`3x*(ln x) = 4x*ln 3 + 4x*ln x`

`3x*(ln x) - 4x*ln x = 4x*ln 3 => -x*(ln x) - 4x*ln 3 = 0`

`x*(ln x)+ 4x*ln 3 = 0 => x*(ln x + 4ln 3) = 0 => {(x=0),(ln x - 4ln 3 = 0):}`

`ln x - 4x*ln 3 = 0 => ln x =4ln 3 => ln x = ln 3^4 => x = 3^4 => x =81`

Notice that the value x = 0 makes the equation `ln (x^(3x)) = ln ((3x)^(4x))` invalid.

**Hence, evaluating the solution to the given equation yields `x = 81` .**

**If x^3x = (3x)^4x, then find the value of x.**

**There's a lot going on here. Firs, let's use an exponent rule:**

**x^3x = (x^3)^x and**

**(3x)^4x = ((3x)^4)^x**

**So we have:**

**(x^3)^x = ((3x)^4)^x**

**Take the log of both sides (remembering log rules):**

**x log x^3 = x log (3x)^4 ( log rules - email for more info on this if it doesn't make sense)**

**Divide both sides by x log:**

**x^3 = (3x)^4**

**x^3 = (3^4)(x^4) (more exponent rules)**

**x^3 = (81)(x^4)**

**(x^3)/(x^4) = 81 (divided both sides by x^4)**

**1/x = 81 (even more exponent rules)**

**x = 1/81**

x^3x = (3x)^4x

x=81

Solve it by any easier way not by logarithm.