Question

solve for x.
a) 9$9^{3x+4}$ = $27^{4x+3}$
b) $4^{3x}$ = $2^{x+1}$

(include steps pleez)

Answer 1

Answer:

Question a)

$9^{3x+4}=27^{4x+3}$

$x=-0.166$

Question b)

$4^{3x}=2^{x+1}$

$x=0.2$

Step-by-step explanation:

Question a)

Given the expression

$9^{3x+4}=27^{4x+3}$

Taking log on both sides

$log\left(9^{3x+4}\right)=log\left(27^{4x+3}\right)$

$\left(3x+4\right)\cdot \left(log\left(9\right)\right)\:=\left(4x+3\right)\cdot \left(log\left(27\right)\right)$

$3x+4=\left(\frac{log\left(27\right)}{log\left(9\right)}\right)\cdot \left(4x+3\right)$

$3x+4=1.5\times \left(4x+3\right)$

$3x+4=6x+4.5$

$3x=6x+0.5$

$\frac{-3x}{-3}=\frac{0.5}{-3}$

$x=-0.166$

Therefore, the value of x:

$x=-0.166$

Question b)

Similarly, we can solve the 'b' expression

Given the expression

$4^{3x}=2^{x+1}$

Taking log on both sides

$log\left(4^{3x}\right)=log\left(2^{x+1}\right)$

$3x\cdot \left(log\left(4\right)\right)=\left(x+1\right)\cdot \left(log\left(2\right)\right)$

$3x=\left(\frac{log\left(2\right)}{log\left(4\right)}\right)\cdot \left(x+1\right)$

$3x=0.5\times \left(x+1\right)$

$3x=0.5x+0.5$

$2.5x=0.5$

Divide both sides by 2.5

$\frac{2.5x}{2.5}=\frac{0.5}{2.5}$

$x=0.2$

Therefore, the value of x = 0.2