Question 169148
Solve algebraically:
1) {{{Log(x+1)+Log(x-1) = Log(3)}}} Apply the "product rule" ({{{Log[b](M)+Log[b](N) = Log[b](M*N)}}}) for logarithms to the left side.
{{{Log((x+1)(x-1)) = Log(3)}}} Simplify the left side.
{{{Log(x^2-1) = Log(3)}}} Apply the identity: If{{{Log[b](M) = Log[b](N)}}}, then {{{M = N}}}, so...
{{{x^2-1 = 3}}} Add 1 to both sides.
{{{x^2 = 4}}} Take the square root of both sides.
{{{highlight(x = 2)}}} or {{{x = -2}}} Discard the negative solution as the log of a negative number is not a real number.
2) {{{5e^(x+2) = 20}}} Divide both sides by 5.
{{{e^x = 4}}} Take the natural log of both sides.
{{{ln(e^x) = ln(4)}}} Apply the "power rule" for logs: {{{Log[b](M^a) = a*Log[b](M)}}}
{{{x*ln(e) = ln(4)}}} Substitute {{{ln(e) = 1}}}
{{{x = ln(4)}}}
{{{x = 1.386}}}
3) {{{4e^x = 48}}} Divide both sides by 4.
{{{e^x = 12}}} Take the natural log of both sides.
{{{x*ln(e) = ln(12)}}}
{{{x = ln(12)}}}
{{{highlight(x = 2.485)}}}
4) {{{9.5e^(0.005x) = 19}}} Divide both sides by 9.5
{{{e^(0.005x) = 2}}} Take the natural log of both sides.
{{{0.005*ln(e^x) = ln(2)}}} Divide both sides by 0.005
{{{x = ln(2)}}}
{{{highlight(x = 0.693)}}}
5) {{{5*Log[3](x) = 10}}} Divide both sides by 5.
{{{Log(x) = 2}}} Rewrite in exponential form: {{{Log[b](x) = y}}} means {{{b^y = x}}} so...
{{{3^2 = x}}}
{{{highlight(x = 9)}}}