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sophiagarces32
How would I solve for x in the equation: 0.406779661=-.188ln(x)+2.6979?
isolate ln(x), so it should look like: ln(x) = (.406-2.6979)/-.188 and then to cancel out the natural log, have both sides be the exponent of e e^ln(x) = e^[(.406-2.6979)/-.188] so x = e^[(.406-2.6979)/-.188].