Solve for x using logarithms. 3^(2-3x) = 4^(2x+1)

Question

asked Sep 19, 2024 220k views

1 Answer

Arjang · Answer 1 · 2024-09-24T20:13:08+0000

Answer:

$\large\boxed{\sf 0.133 }$

Explanation:

The given equation to us is ,

$\longrightarrow 3^(2-3x)= 4^(2x+1) \\$

Take log to base 10 on both sides,

$\longrightarrow \log_(10) 3^(2-3x)= \log_(10) 4^(2x+1) \\$

Now recall that,
$\log a^m = m\ log \ a$

$\longrightarrow (2-3x) \log_(10) 3 = (2x+1)(\log_(10) 4)\\$

We can write it as ,

$\longrightarrow (2-3x) \log_(10)3=(2x+1)\log_(10)2^2 \\$

Again using the property mentioned above,

$\longrightarrow (2-3x)\log_(10)3 = 2(2x+1)\log_(10) 2 \\$

$\longrightarrow (2-3x)\log_(10)3 = (4x+2)\log_(10) 2 \\$

Now we know that ,

On substituting the values, we have ,

$\longrightarrow (2-3x)0.477 = (4x+2)0.301 \\$

simplify by opening the brackets,

$\longrightarrow 0.954 - 1.431x = 1.204x + 0.602 \\$

$\longrightarrow 1.204x + 1.431x = 0.954 - 0.602 \\$

$\longrightarrow 2.635x = 0.352 \\$

$\longrightarrow x =(0.352)/(2.635) \\$

$\longrightarrow \underline{\underline{ x = 0.133 }} \\$

and we are done!