How do you solve $4 (7^{x + 2}) = 9^{2 x - 3}$ $4(7^(x + 2)) = 9^(2x - 3)$ ?

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Answer 1 · 2018-04-22T14:28:43

$x = \frac{- 3 ln (9) - 2 ln (7) - ln (4)}{ln (7) - 2 ln (9)}$ $x=(-3ln(9)-2ln(7)-ln(4))/(ln(7)-2ln(9))$

you have to log the equations

$4 \cdot 7^{x + 2} = 9^{2 x - 3}$ $4*7^(x+2)=9^(2x-3)$

Use either natural logs or normal logs $ln$ $ln$ or $log$ $log$ and log both sides

$ln (4 \cdot 7^{x + 2}) = ln (9^{2 x - 3})$ $ln(4*7^(x+2))=ln(9^(2x-3))$

First use the log rule that states $log a \cdot b = log a + log b$ $loga*b=loga+logb$

$ln (4) + ln (7^{x + 2}) = ln (9^{2 x - 3})$ $ln(4)+ln(7^(x+2))=ln(9^(2x-3))$

Remember the log rule that states ${log x}^{4} = 4 log x$ $logx^4=4logx$

$ln (4) + (x + 2) ln (7) = (2 x - 3) ln (9)$ $ln(4)+(x+2)ln(7)=(2x-3)ln(9)$

$ln (4) + x ln (7) + 2 ln (7) = 2 x ln (9) - 3 ln (9)$ $ln(4)+xln(7)+2ln(7)=2xln(9)-3ln(9)$

Bring all the $x ln$ $xln$ terms to one side

$x ln (7) - 2 x ln (9) = - 3 ln (9) - 2 ln (7) - ln (4)$ $xln(7)-2xln(9)=-3ln(9)-2ln(7)-ln(4)$

Factorise the x out

$x (ln (7) - 2 ln (9)) = (- 3 ln (9) - 2 ln (7) - ln (4))$ $x(ln(7)-2ln(9))=(-3ln(9)-2ln(7)-ln(4))$

$x = \frac{- 3 ln (9) - 2 ln (7) - ln (4)}{ln (7) - 2 ln (9)}$ $x=(-3ln(9)-2ln(7)-ln(4))/(ln(7)-2ln(9))$

Solve on the calculator using the ln button or if your calculator doesn't have it use the log base 10 button.

How do you solve 4(7^(x + 2)) = 9^(2x - 3)4(7x+2)=92x−34(7^(x + 2)) = 9^(2x - 3)?