Question 295377
1) Solve for a:
{{{Log[10](a)+Log[10](a+21) = 2}}} Apply the product rule for logarithms:{{{Log[b](M)+Log[b](N) = Log[b](M*N)}}}
{{{Log[10](a)+Log[10](a+21) = Log[10](a*(a+21))}}} Simplify the argument on the right side.
{{{Log[10](a^2+21a) = 2}}} Rewrite this in exponential form.
{{{10^2 = a^2+21a}}} Rearrange into the standard quadratic equation form.
{{{a^2+21a-100 = 0}}} Solve by factoring.
{{{(a-4)(a+25) = 0}}} Apply the zero product rule.
{{{a-4 = 0}}} or {{{a+25 = 0}}} so then...
{{{a = 4}}} or {{{a = -25}}} Discard the negative solution*.
{{{highlight(a = 4)}}}
Check: Substitute a = 4.
{{{Log[10](a)+Log[10](a+21) = Log[10](4)+Log[10](25)}}} = {{{2}}}
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2) Solve for a.
{{{Log[6](a^2+2)+Log[6](2) = 2}}} Apply the product rule.
{{{Log[6](2*(a^2+2)) = 2}}} Rewrite this in exponential form.
{{{6^2 = 2a^2+4}}} Subtract 4 from both sides.
{{{32 = 2a^2}}} Divide both sides by 2.
{{{a^2 = 16}}} Take the square root of both sides.
{{{a = 4}}} or {{{a = -4}}} Discard the negative solution*.
{{{highlight(a = 4)}}}
*The logarithm of a negative quantity yields a complex number.