# If log7 (a) + loga (7) = 2, what is a?

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We have log (7) a + log (a) 7 = 2

We use the relation for logarithms that log a + log b = log (a*b) and log (a) b = 1 / log(b) a.

log (7) a + log (a) 7 = 2

=> log (7) a + 1/ log (7) a = 2

=> [log (7) a]^2 + 1 = 2* log(7) a

=> [log (7) a]^2 + - 2* log(7) a + 1 = 0

=> (log(7) a - 1)^2 = 0

=> log(7) a = 1

=> a = 7

**The required solution is a = 7.**

log7 (a) + loga ( 7) = 2

We will rewrite using the logarithm properties.

We know that:

log7 (a) = loga (a) / loga (7)

But loga a = 1

==> log7 (a) = 1/loga (7)

Let us substitute into the equation.

==> 1/loga (7) +loga (7) = 2

==> Multiply by loga (7)

==> 1 + ( loga (7))^2 = 2loga (7)

Let us assume that loga (7) = x

==> 1 + x^2 = 2x

==> x^2- 2x +1 = 0

==> (x-1)^2 = 0

==> x = 1

==> loga (7) = 1

**==> a = 7**