$\displaystyle log(2x+5) * log(9x^2) = 0$ Thanks a whole bunch!
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Originally Posted by hydride $\displaystyle log(2x+5) * log(9x^2) = 0$ Thanks a whole bunch! either $\displaystyle 2x+5 = 1$ or $\displaystyle 9x^2 = 1$
Originally Posted by hydride $\displaystyle log(2x+5) * log(9x^2) = 0$ Thanks a whole bunch! I'll assume that the base of your logarithm is 10, however it doesn't really matter. In the case that [tex]log_{10}(2x+5)=0, we have $\displaystyle 10^0=1=2x+5$. Just solve this equation for x. Do the same with the other factor.
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