J jack_r12 New member Joined Feb 16, 2017 Messages 1 Feb 16, 2017 #1 Could someone help me solve the equation, (18^2x)(3^-2x)=6 I'm assuming I'll use logs somewhere in that equation Last edited by a moderator: Feb 17, 2017
Could someone help me solve the equation, (18^2x)(3^-2x)=6 I'm assuming I'll use logs somewhere in that equation
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 Feb 16, 2017 #2 jack_r12 said: Could someone help me solve the equation, (18^2x)(3^-2x)=6 I'm assuming I'll use logs somewhere in that equation Click to expand... To begin, let's rewrite the equation as follows: \(\displaystyle \displaystyle \frac{18^{2x}}{3^{2x}}=6\) Now, applying the rule for exponents \(\displaystyle \displaystyle \frac{a^{c}}{b^{c}}=\left(\frac{a}{b}\right)^c\), what do you get?
jack_r12 said: Could someone help me solve the equation, (18^2x)(3^-2x)=6 I'm assuming I'll use logs somewhere in that equation Click to expand... To begin, let's rewrite the equation as follows: \(\displaystyle \displaystyle \frac{18^{2x}}{3^{2x}}=6\) Now, applying the rule for exponents \(\displaystyle \displaystyle \frac{a^{c}}{b^{c}}=\left(\frac{a}{b}\right)^c\), what do you get?