Solve each of the following. (Round answers to 3 decimal places) 1) 3^2x-4=5 3^5x+6=4^x 3) 5^2x-4=3^4x-5 4) (1+(0.10)/(12))^12t=2 5) lnx+ln(x-2)=1 6) e^2x=7 7) 2-6ln3x=10 8) log_(2)x+log_(2)(x+2)=log_(2)(x+6) 9) e^2x+3e^x-10=0

1) $3^{2x-4}=5$<br />Taking the logarithm of both sides, we get:<br />$(2x-4)\ln(3)=\ln(5)$<br />Solving for $x$, we get:<br />$x=\frac{\ln(5)+4\ln(3)}{2\ln(3)}\approx 1.464$<br /><br />2) $3^{5x+6}=4^{x}$<br />Taking the logarithm of both sides, we get:<br />$(5x+6)\ln(3)=x\ln(4)$<br />Solving for $x$, we get:<br />$x=\frac{6\ln(3)-\ln(4)}{5\ln(3)-\ln(4)}\approx 0.643$<br /><br />3) $5^{2x-4}=3^{4x-5}$<br />Taking the logarithm of both sides, we get:<br />$(2x-4)\ln(5)=(4x-5)\ln(3)$<br />Solving for $x$, we get:<br />$x=\frac{4\ln(3)+5\ln(5)}{2\ln(5)-4\ln(3)}\approx 1.732$<br /><br />4) $(1+\frac {0.10}{12})^{12t}=2$<br />Taking the logarithm of both sides, we get:<br />$12t\ln(1+\frac{0.10}{12})=\ln(2)$<br />Solving for $t$, we get:<br />$t=\frac{\ln(2)}{12\ln(1+\frac{0.10}{12})}\approx 5.776$<br /><br />5) $lnx+ln(x-2)=1$<br />Using the properties of logarithms, we can combine the two logarithms:<br />$\ln(x(x-2))=1$<br />Exponentiating both sides, we get:<br />$x(x-2)=e$<br />Solving for $x$, we get:<br />$x=\frac{e+2}{2}\approx 3.718$<br /><br />6) $e^{2x}=7$<br />Taking the natural logarithm of both sides, we get:<br />$2x=\ln(7)$<br />Solving for $x$, we get:<br />$x=\frac{\ln(7)}{2}\approx 1.099$<br /><br />7) $2-6ln3x=10$<br />Solving for $x$, we get:<br />$ln(3x)=-\frac{8}{6}$<br />Exponentiating both sides, we get:<br />$3x=e^{-\frac{8}{6}}$<br />Solving for $x$, we get:<br />$x=\frac{e^{-\frac{8}{6}}}{3}\approx 0.183$<br /><br />8) $log_{2}x+log_{2}(x+2)=log_{2}(x+6)$<br />Using the properties of logarithms, we can combine the three logarithms:<br />$log_{2}(x(x+2))=log_{2}(x+6)$<br />Since the bases are the same, we can equate the arguments:<br />$x(x+2)=x+6$<br />Solving for $x$, we get:<br />$x^{2}+2x-x-6=0$<br />$x^{2}+x-6=0$<br />Factoring, we get:<br />$(x+3)(x-2)=0$<br />So, $x=-3$ or $x=2$<br />Since $x=-3$ is not in the domain of the original equation, the solution is $x=2$.<br /><br />9) $e^{2x}+3e^{x}-10=0$<br />Let $u=e^{x}$, then the equation becomes:<br />$u^{2}+3u-10=0$<br />Factoring, we get:<br />$(u+5)(u-2)=0$<br />So, $u=-5$ or $u=2$<br />Since $u=e^{x}$ cannot be negative, the solution is $u=2$, which means $e^{x}=2$<br />Taking the natural logarithm of both sides, we get:<br />$x=\ln(2)\approx 0.693$