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8e^3x+9=13 round to the nearest tenth? its logarithims -_-....
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\[8e^{3x}=4\]
\[e^{3x}=\frac{1}{2}\]
oops i meant thousandths.... and it has to be in decimal form.
now what can you do to get the e to cancel
divide it by itself??
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\[\ln(e^x)=x\]
that is basically all you need to remember whe you want to get rid of an e raised to variable
so if you multiply both sides by ln
\[\ln(e^{3x})=\ln(\frac{1}{2})\] \[3x=\ln(\frac{1}{2})\]
\[\ln(\frac{a}{b})=\ln(a)-\ln(b) \] this is the next rule you'll need
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\[3x=\ln(1)-\ln(2)\] \[x=\frac{\ln(1)}{3}-\frac{\ln(2)}{3}\]
\[x=\frac{\ln(1)-\ln(2)}{3}\]
-0.231?
if thats what you got then yes
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