Question

how so you slove 4^2x-1 = 8^x+2 ?

Answer 1

Answer:

x = $\frac{ln(16)}{ln(2)}$

Step-by-step explanation:

$4^{2x-1} =8^{x+2}$

ln both sides like you would add the same number to both sides to keep the equation equivalent

$ln4^{2x-1} = ln 8^{x+2}$

using log rules, exponents can be brought down as a constant multiplying ln

$(2x-1)ln(4) =(x+2)ln(8)$

use distributive property to remove parentheses

2xln(4) - ln(4) = xln(8) + 2ln(8)

isolate x

2xln(4) - xln(8) = 2ln(8) + ln(4)

rewrite the logs using log rules: ln(x) - ln(y) = ln($\frac{x}{y}$) and the exponent rule stated above

xln($4^{2}$)-xln(8) = ln($\frac{8^{2} }{4}$)

factor out x

x(ln$4^{2}$-ln8) = ln($\frac{8^{2} }{4}$)

again use log rules:  ln(x) - ln(y) = ln($\frac{x}{y}$)

x* $ln(\frac{4^{2} }{8}$$)$=  ln($\frac{8^{2} }{4}$)

now simplify the stuff inside the ln parentheses

x*ln(2) = ln(16)

x = $\frac{ln(16)}{ln(2)}$